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y-fmen/ of Educailon 




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SMITHSONIAN' DKI'OSIT. 



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PEESPECTIYE 



AND 



GEOMETEICAL 



DRAWING 



ADAPTED TO THE USE OF CANDIDATES FOB 



SECOND AND THIRD-CLASS TEACHERS' CERTIFICATES, 



BY 
TTHOS. H. IVtcaUIRIv, B.A. 

CoiU(E££;iAL Master, CoLLUsfawooD Collsgiatb Instititts. 



AUTHORIZED BY THE DEPARTMENT OF EDUCATION. 



TORONTO: 



WILLIAIvI BRIQQS, 

WESLEY BUILDINGS. 
C. W. COAXES. Montreal. Qub. S. F. HUESTIS, Halifax, N.a 



FEB QSVo^^ 



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K 



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Entered according to Act of the Parliament of Canada, in the yea? one thousand 
eight hundred and eighty-seven, by William Briggs, Book Steward of the 
Methodist Book and Publishing Etouse, Tarcmto, ^t the Department at 
Agriculture. 



PEEFACE. 



Drawing having been at length recognized by the Educa- 
tion Department as an essential feature in High School edu- 
cation, it is necessary that a work, at once simple and concise, 
should be prepared on this subject. The incompleteness and 
want of definiteness in the existing works on perspective, have 
induced me to place this book before High School pupils. It 
consists mainly of problems, etc., that have been given from 
time to time in my own classes. To obviate the necessity of 
copying problems from the blackboard, I have added a num- 
ber in Geometrical Drawing, which will be found useful. 

Believing that this work supplies a want long felt in our 
schools, I have consented to place it before the public 

T. H. M. 

The Institute, MarcK 1887. 



CONTENTS. 



PAOE. 

Introduction 9 

Drawing to a Scale 15 

The Point 16 

Exercise on the Point 22 

The Line 23 

Exercise on the Line , 27 

Surfaces — The Square 28 

Exercise on the Square 36 

The Oblong 37 

Exercise on the Oblong . . 38 

The Triangle 39 

Exercise on the Triangle 45 

The Hexagon 47 

Exercise on the Hexagon 52 

The Octagon 54 

The Circle 67 

Exercise on the Circle 63 

Solids 64 

The Cube 66 

The Plinth 68 

Exercise on the Plinth 70 

The Prism 72 

The Cylinder 75 

The Pyramid 78 

The Cone 80 

Exercise on the Prism and Cylinder 81 

Exercise on the Pyramid and Cone 82 

The Frusta 83 

Exercise on the Frusta 85, 87 



, VI CONTENTS. 

Solids — Continued. page. 

The Sphere 88 

Exercise on the Sphere . 90 

Foreshortening -92 

Synthetic Perspective , 93 

Perspective Effect 94 

Angular Perspective — Figures on Picture Plane 96 

Figures within the Picture Plane 99 

Exercise in Angular Perspective 101 

Miscellaneous Exercises 101 

Geometrical Drawing 105 

To draw a perpendicular to a given line from a point on the 

line or away from it 105 

To describe a square on a given straight line 106 

To describe a square on a given diagonal 107 

To construct an oblong of given dimensions 107 

To bisect a given line 107 

To divide a given line into any number of equal parts 108 

To draw a line parallel to a given line from an external point . 108 

To divide a line proportionally to another divided line 109 

To construct a triangle of given dimensions 109 

To bisect a given angle 110 

To trisect a right angle 110 

To inscribe a circle in a given triangle Ill 

To draw a circle through three given points Ill 

To find the centre of a given circle or arc 112 

To draw a tangent to a circle from a point without or on the 

circumference 113 

To draw an isosceles triangle of given dimensions 114 

To draw an equilateral triangle of given dimensions 114 

To draw from a point an angle equal to a given angle 115 

To construct a triangle similar to a given triangle within a 

given circle 115 

To construct an equilateral triangle about a given circle 116 

To construct a triangle similar to a given triangle about a 

given circle 117 

Within a circle to draw any number of equal circles, each 

touching two others and the outer circle 118 

To construct a polygon on a given line 119 



CONTENTS. Vll 

GEOMHtlEitCAri BraWIN^g — Continmd. paqi&. 

To construct a regular polygon in a given circle %......»** v » . 120 

To construct a regular pentagon oxi a given line . . . ^ ^ * 120 

To construct a regular hexagon on ^ given line ***•..»... 121 

To construct a regular octagon on a given line ,».*.. 122 

To construct a regular octagon in a given squatie ..»..*» 122 

To construct an ellipse with axes and intersecting arcs , 123 

To construct an ellipse with concentric <3ircles , w . * • 124 

To construct an ellipse when the longer axis only Is given . * *. . 125 

To find the axes and foci of a given ellipse w .... * > . 125 

To draw a tangent to an ellipse from a point in the curvie .... 126 

To draw a tangent to an ellipse from an external point * . 127 

To draw an oval of a given width 127 

To draw an involute to a circle 128 

To draw a mean proportional (greater) to two given straight 

lines 1^29 

To draw a mean proportional (less) to two given straight 

lines 130 

To draw a circle touching two lines not parallel » 130 

To draw a circle touching another circle and a given line 131 

Graded Exercise on Geometrical Drawing 132 



DEAWING, 



<ri 



RAWING may be defined as the representation of an 
J\^-^ object or collection of objects on a plain or level 
surface. 

There are two kinds of Drawing : Perspective, or the 
representation of an object as it appears to the eye ; and 
Geometrical Drawing, as it actually is. 

Our knowledge of the size of an object can be known only 
by experiment. We must either see the object near the eye, 
and observe its size, or know its distance from the eye. 

From long practice we are enabled to tell the height of a 
hill, breadth of a river, or capacity of a ship, though at a con- 
siderable distance. If we hold a rule of definite length close 
to the eye, and then withdraw it six or eight feet away, we 
notice that it is apparently smaller. Experience teaches us 
that it is not really smaller, the apparent diminution being 
only the effect of distance. 

Perspective aims, then, at measuring and representing 
objects as they appear at a distance. 

The horizon always hounds our vision. 

If we look out upon a large lake we find that the sky and 
water appear to meet ; this line of apparent union is called 



10 



DRAWING. 



the horizon. No matter how large an object is, if it recedes 
far enough from us on a lake, it would at length appear 
on the horizon as a point. 

If a person stands on a level plain he can see over a range 
of 60 degrees in every direction without moving his head. 
The point on the horizon directly in front of him is called 
the Centre of Vision. 







Now, if we take this point as a centre, and join it with the 
eye of the observer, and draw lines at an angle of 30° with it 
from the observer's eye, we shall form a hollow cone, which 
represents his range of vision : thus, if the observer's eye be 
at the point X, and AB represent the horizon, and the 
point on it directly in front of X, if CX be joined, and 
AX, BX be drawn at an angle of 30 degrees with CX, AB 



INTEODUCTION. 



11 



will be the horizontal range of the spectator's vision. If we 
describe a circle from centre C, and distance CA or CB, such 
circle will be called the base of the cone of visual rays ; for it 
will be observed that from point X the spectator can see just 
as far as the edge of this circle. The ground as a plane is 
generally supposed to extend to the horizon, and is marked 
for sake of abbreviation, G. P. The horizontal line is un- 
limited in length, and is written H. L. (Fig. l.^ 




All representations of objects are supposed to be made on a 
plane (unlimited in extent), perpendicular to, and resting on, 
the ground plane, and directly in front of the spectator. This 
plane is called the Picture Plane^ and is written P. P. 

It must be distinctly borne in mind that the bottom of the 
picture plane touches the ground plane, and this line is always 
at right angles to that joining the spectator's eye with the 
centre of vision. 



12 DRAWINa 

The line joining the spectator's eye with the centre of 
vision is usually denoted L. D. (line of direction, and some- 
times length of distance). 

The position of the spectator's eye is called the point of 
station, or P. S. (Fig. 2.) 

If a spectator stand at O and observe a stick at AA, then 
carry th^ stick back and place it at XX, parallel to AA, and 
draw lines OX, OX ; it is plain the stick cannot appear as 
large as AA, but will be represented in length by BB. If 
again withdrawn to YY, it will appear as ee, and so on. 
As we remove it the apparent length diminishes, or, in 
other words, the angle made by the line with O constantly 
decreases. (Fig. 2.) 

If the stick were removed to an infinite distance, its appar- 
ent length would vanish to a point S, and the lines drawn to 
O would coincide, forming an angle of O degrees. 

Objects appear to diminish in size as they recede from the 
eye, and vice versa. 

At the centre of vision all objects have no apparent magni- 
tude, and must be represented by a point, and steadily in- 
creased in size till they approach the eye. This is why rails 
on a straight piece of railway appear to meet in the distance, 
though everywhere the distance between them is the same. 

Now, if we stood on a straight piece of track, and if other 
tracks were laid on each side and parallel with it, every tracl' 
would appear to vanish at the same point directly in front of 
us, hence the rule : 

Lines parallel to the line of direction will vanish at the 
centre of vision. (Fig. 3.) 

Thus, let C be centre of vision, and AA, BB the ends of 
parallel lines : they will all meet at C, and all the lines drawn 
from C to AB will really be right angles, however different 
they may appear to be. Now draw DD parallel to AA, and 



INTRODUCTION. 



13 



since the distance from A to A is always the same as from D 
to D, we have the rule : 




Parallel lines drawn between vanishing lines are of equal 
length, and, conversely, the lines joining the extremities of 
equal parallel lines will vanish to a 'point. 




Fig. 4. 



Again, if the spectator stood on the ground at O, where the * 
picture plane rests, and looked towards C on tlie horizon, it 



14 DKAWING. 

is clear that he could see objects lying on the ground any- 
where between AB and HL. Hence, if we take AB as the 
ground line, or base of picture plane, and OC as the height of 
spectator's eye above the ground, the horizontal line will pass 
through C and be parallel to AB. (Fig. 4.) 

The ground line is denoted G. L. 

If a stake AS be placed perpendicularly in the ground 
plane at A, it will necessarily touch P.P. throughout its length. 
If placed as TT, still erect, it is said to be within the picture 
plane, and parallel to it. Now, if the stake be same height 
throughout, TT will be equal to AS ; so also will YY be equal 
to AS ; and being placed on the ground, their extremities will 
lie in the straight lines SY, AY (being parallel). But lines 
joining equal parallel lines will vanish, hence SY and A^", 
being produced, will meet in C. 

As nothing is ever supposed to be drawn nearer to the eye 
than the picture plane, we use the picture plane as a basis of 
measurement, — for in the figure we can estimate the length 
of TT or YY only by referring them to AS, which is drawn 
on the picture plane. 

Objects to the right of OC are said to be to the right; 
objects left of OC, to the left. Objects which are at an equal 
distance on each side are said to be directly in front. 

Perspective is of four kinds : — 

1. Parallel, in which some side or face of the object is per- 
pendicular or parallel to the P.P., and also to the G.P. 

2. Angular, in which a side or face makes an angle less 
than a right angle with the P.P., but is parallel to the G.P. 

3. Oblique, jn which the sides or faces make angles less 
than right angles with both P.P. and G.P. 

4. Aerial, or the perspective of distinctness in a view. It 
• is related to shading and painting. 



THE POINT. 



15 



DEAWING TO A SCALE. 

Unless objects were very small, and our drawing surface 
large, we could not represent the size of the object as it is. 
It is usual to draw the figure in miniature, or a certain num- 
ber of times smaller. Every line in the drawing must bear 
this fixed proportion to the corresponding one in the figure or 
object. This is called drawing to a scale. 

Usually, one quarter-inch for each foot is the proportion in 
perspective, but any other ratio may be used. 

Thus, if a line eight feet in length were to be drawn, it 
would be represented by a line (on the picture plane) two 
inches in length. 



HI 


i 








HI 




d 
d 


\ 


V\6 


\ 








\.\ 




EL 




\ 


\ BL, 




D 




A 




B 



Fig. 5. 

Draw any norizontal line H.L., take any point O, draw OC 
perpendicular to H.L. Let CD = height of spectator; through 
D draw B.L. parallel to H.L. If O be spectator, C will be 
centre of vision, B.L. base line, and H.L. horizontal line. 00 
will be line of direction and OD length of distance. (Fig. 5.) 



16 DRAWING. 

In B.L. take any points, A, B, and join them with centre 
of vision (C.Y.). If dah be drawn parallel to DAB, and 
touching CD and CB, it will be equal to DB, and da = DA 
and ab = AB ; also, if the distance of B to the right of D be 
known, the distance of b from d is known, for db = DB ; or, in 
other words, every part of BC is the game distance from CD 
that B is. 

If we wish to find the position of a point within the plane, 
we first find its distance on B.L. from D, and then join the 
point marking this distance with C; the latter line would pass 
through it. 

Exercise I. 

1. Find a point on the ground plane, at base of picture 
plane, 4 feet to right. 

2. Find a point 3 feet above G.P., 4 feet to right, and 
touching P.P. 

3. Find a point directly in front touching P.P., and 6 feet 
above it. 

4. Find position of a point 3 feet under G.P., 3 feet to 
right in P.P. produced. 

(In foregoing examples assume spectator's height to be six 
feet.) 

All lines at Hght angles with the base line (B.L.)., or, 
parallel with the direction we are looking (L.D) will vanish 
at the centre of vision (C.Y.); but lines parallel with the 
base line (B.L.) or horizon (H.L.) will never vanish, but 
always appear parallel. This is an important rule. 



To find the distance of a point within the plane. 

In fig. 6, let AB DC represent the face of a cube lestmg 

on the ground, and let the given face touch the picture plane. 

The near lower edge will then coincide with base line AB. 

Now, if we suppose the cube placed to the left of the spec- 



THE POINT. 



17 



tator, he will be able to see the side BD FE, or the top, if he 
3 as 

HL 



he as high as the cube. 




We have said that all lines at right angles to AB will 
vanish at point C. Y. ; and since all angles of a cube are right 



18 DRAWING. 

angles, the line BE, if produced, will reach C.V., and 
so will line DF, for the same reason. But we want to find 
the position of E, the extremity of the line BE. 

Draw a line from C.Y. perpendicular to AB, and produce 
it indefinitely. A spectator stationed at N would see the 
edge BE represented in size by the dotted line Be perpen- 
dicular from B on EN. If he were placed at O, BE would 
appear as B6, perpendicular to EG from B ; also, if placed at 
P, BE would appear as Ba / and it will be noticed that, as 
the spectator recedes, the apparent size of BE decreases, ?*.e., 
Ba is less than B6, and B6 less than Be, and so on. (Fig. 6.) 

Now from B measure lengths of Be, B^, Ba on BE : thus, 
B/= Be, Be = Bb, Bd = Ba, etc. 

Take a point M on base line at a distance from B equal to 
required length of BE ; this point may be on either side of B. 

In given case, since BE = AB or BD, make BM = AB, and 
suppose M to be drawn to right of B. Join M/, Me, MJ, etc., 
and produce them backward to horizontal line touching it in 
h, k, m, etc. 

Now, if M be a fixed point for a distance from B, it will be 
seen that as the spectator recedes from the object the points 
h, k, m, etc., recede from C.Y. ; hence, if the points ?i, A, m, 
etc., be given, and knowing position of M, we can find appar- 
ent distance of BM within the plane, as shown by B/, Be, etc., 
and this is done by the following method : 

Take C.Y. as centre, and distances N, O, P, etc., of specta- 
tors from object, as distance; and describe a semicircle cutting 
horizontal line in m, k, or h on one side, and corresponding 
points on the other. The points where the semicircle cuts 
the horizontal line, are called the measuring points, and are 
denoted by RMP, LMP according as they are to the right or 
left of C Y 



THE POINT. 



19 



To find the measuring points, having given the height of 

spectator and his distance from the picture plane or base line. 

In fig. 7, let II.L. be horizontal line, which is unlimited in 



a. 


r 




^ 


V 


^ 


\ 




O 


^ 


t> 


caV 


A^ 


Q 

* 


07 


^ 


/ 






/ 




L-- 


s 










length ; take any point C. V. in it, and draw line L.D. from it 
at right angles. Make C. Y. = 6 feet, or whatever may be 
the spectator's height, and through O draw B.L. parallel to 
H.L.; produce C.V. O to S.P., so that O S.P. equals distance 



20 DRAWING. 

of spectator from base line. Then with centre C.Y. and 
distance C.Y. S.P. describe a semicircle, "cutting H.L. in LMP 
and RMP, the right and left measuring points respectively. 

In Hg 7, let any point A be taken, say 4 feet to the left, on 
the base line ; join A C.Y. 

Suppose we wish to find a point the same distance to the 
left that A is, but 4 feet within the plane : we know that the 
point lies somewhere on A C.Y., because every part of this 
line is the same distance from C.Y. O that A is. Now, we 
proceed by measuring the required distance to right or left of 
A, and joining the point thus found with the measuring point 
opposite — that is, if point be taken to the right of A, as O, 
we join O LMP ; if to the left, as E, we join E KMP. O 
LMP and E RMP will always cut A C.Y. at the same point 
Bif EA = AO. 

So also, if we take a lesser distance, as AD, and make 
AE = AD; join F RMP and D LMP; they will intersect in c, 
Now AO or AE = AB, hence B is four feet within the plane 
and four feet to the left. 

In practice it is not necessary to draw to both measuring 
points, one (the nearest) will answer every purpose. 

Example 1, — Find position of a point on the ground 6 feet 
directly in front, as seen by a spectator 6 feet in height, and 
4 feet from picture plane. 

In fig. 8, we draw H.L. and B.L. 6 feet apart, and C.Y. 
S.P. perpendicular to B.L. from C.Y., and make O S.P. 
the spectator's distance from base line. Draw a semicircle 
to cut H.L. in LMP and BMP. Now, since point required 
is on line between C.Y. and O, we measure 6 feet either way 
on B.L., as A or B, and join to measuring point as B LMP 
or A BMP ; they will intersect in X, the point required. 




Example 2. — Find a point 3 feet (3^) to right, 4 feet (4') 
within the P.P., and 5 feet (5^) above it. Height of spectator 
5 feet 6 inches (5' 6''), and his distance from P.P. 4 feet (4'). 



BMP 




Fig. 9. 



Draw H.L. and B.L. 5' 6'' apart; draw C.Y. S.P. cutting 
B.L. in O ; make O S.P. = 4'. Find EMP (C.Y. EMP must 
always be equal to C.Y. S.P.); make 0A = 3', from A measure 



22 DRAWING. 

off AB equal to the required distance of point within the 
plane (4'). Join A C. Y. and B BMP, intersecting in C ; draw 
AD perpendicular to B.L., and make AD equal to required 
height (5^). Join D C.Y. Draw CX parallel to AD and 
meeting D C.Y. in X. X is position required. For, since 
AC = AB, and any point C in A C.Y. is same distance to the 
right that A is, then C is 3' to the right and 4' within P.P. 

Now, D C.Y. and A C.Y. are vanishing lines, and CX and 
AD are parallel lines drawn between them, then CX = AD ; 
but point D is 5' high, then X is the same height, and is 
vertically above C also; therefore X is position of required 
point. (Fig. 9.) 

Note. — ^All measurements must be made on the P.P. 

Exercise II. 

(In the following examples take height of spectator 6 feet, 
and his distance from P.P. 4 feet, and make scale ^ inch to 
one foot.) 

1. Find position of a point directly in front, and 8 feet 
within P.P. on G.P. 

2. Find position of a point 4' to left (L), 4' within P.P. on 
G.P. 

3. Find position of a point 6^ to right (B), 6' within P.P., 
and 4' above G.P. 

4. Find position of a bird flying 10' to B., 12' within P.P., 
and 8' above G.P. 

5. Find position of a fish resting in water 4' beneath G.P., 
6' to B., and 8' within P.P. 

6. In Ex. 1, show that however far the point be away, it 
must always be nearer than the point C.Y. 



THE LINE. 



23 



THE PEESPECTIVE LINE. 

A straight line is the shortest distance between any two 
points. If we know the position of any two points we can 
locate the line between them. 

Example 1. — Draw a staff 8' high placed erect on ground 
plane, 6' to R., and M within P.P.; distance 4', height of 
spectator 6' (H = 6'), scale ^' = T. 



m? 




J'ig. 10. 



Draw H.L. and B.L. 6' apart ; draw C. Y. S.P., making 
O S.P. (LD) 4', and find PMP as before. Take A, 6' to riglit 
of O, and erect perpendicular AB 8' in height; join B C.Y. 



24> 



DRAWING. 



and A C.Y.; take C 4' to left of A, and join EMP, cutting 
A C. Y. in E. Through E draw ED paraUel to AB ; ED is 
line required. Now, since AE = AC, E is 6' to right and 4' 
witliin P.P.; but ED = AB, then ED is 4' within P.P., S' 
high, and 6' to right. (Fig. 10.) 

Example 2. — Draw a line 3' long lying on G.P., parallel to 
P.P. (or base line), and 4' within it^ near extremity of line 
being T to left. H = 6', L.D. = 4^ scale I" = 1'. 



LMP 


HL 




w 




BMP 


•s 






A 


" 






1 \ 




""^^./ 


/ 






1 


\ 


/" 


7 


/ 


BL \ 


, / 


/ 






^^ 


/ 




\ ^ 


[i 




Q 


E 


■^ f 



5P 

Fig.lL 



Draw H.L., B.L. and L.D. as before, and find also LMP. 
From O mark off OB = 2' (distance of near extremity from O), 
and make BA=3' (length of line) ; join B C.Y. and A C.Y. 
From B measure off BE = 4' (distance within Y.Y,)'^ join 
E LMP, cutting B C.Y. in D ; through D draw DC to line 
Ik. C.Y. and parallel to AB. CD is line required, for it is 
equal to AB, and its near extremity D is same distance to 
the left that B is, and BE = BD. (Fig. 11.) 

Example 3. — Draw a line 3' in length lying on G.P., per- 



THE LINE. 



25 



pendicular to P.P. (base line), 4' to R., and nearest extremity 
2' within P.P. H = 6', L.D. = 4', scale \;' = T. 



RMP 




Draw H.L., B.L., L.D., as before, and mark point RMP. 
Make OC = 4', join C C. Y. From C measure off CB = 2^, and 
from B mark off BA=3' (length of line) ; join B RMP and 
A RMP, to cut C C. Y. in E, D. ED is the line required, for 
DE is parallel to LD, and therefore perpendicular to BL. 
EC = BC and DC = AC, then DE = AB, and point E is the 
same distance from L.D. that C is. (Fig. 12.) 

Example Jf. — Draw a line 3^ in length parallel to P.P. and 
4' within it, and parallel to G.P., and 4' above it, line to 
be drawn with near extremity 2' to left. 11 = 6', LD = 4', 
scale :i'' = r. 

In fig. 13 draw H.L., B.L. and L.D., and find LMP, 
take B 2' to left and A 4' to left of B, also S 4' to right of 
B. Erect AC and BD perpendiculars to AB, and each 4' in 
height; join CD, C C.Y, D C.Y., B C.Y., and S LMP. Let 
S LMP cut B C. Y. in G; draw GF parallel to BD, and FE 
3 



26 



DRAWING. 



parallel to CD : then EF is line required. For since BG = 
BS, and BD = FG, then DF = BG, so also CE = DF. But 
EF = CD, then EF is parallel to P.P., is same height above 
G.P. as BD, and the same distance to the left as BD. 




THE LINE. 27 

Exercise III. 

(H = 6^ L.D. = 4^ scale Y = I'O 

1. Draw a line 4' in length on G.P., parallel to P.P. and 4' 
within it, directly in front. 

2. Draw a line 4' in length parallel to G.P. and 4' above it, 
touching P.P. 4:' to left, and perpendicular to it. 

3. Draw a line 3' in length perpendicular to G.P., and 3' 
above it ; line to be 4' to right. 

4. Draw a line i' in length parallel to G.P. and 5' above it, 
and in contact with P.P. ; line to have one extremity 3' to 
right. 

5. A xme 5' in length is drawn parallel to P.P. and 6' 
within it ; it is parallel to the G.P., one extremity being 3' to 
right, the other 2' to left. It is 4' above the G.P, 



28 DRAWING. 



SUEFACES IN PEESPECTIYE. 



RECTANGULAR SURFACES. 



Xh.e Square. 

A square is a parallelogram having two adjacent sides 
equal, and the included angle a right angle. 

We said previously that an object appears to decrease in 
size as it recedes from the eye. If, therefore, a square is 
placed on the ground plane with one side touching the picture 
plane, it is clear that the side most removed will appear 
smaller than that touching the picture plane, and so the 
square may not appear to have even one right angle. 

Let AB, BC, CD, DE be all taken of equal length, and let 
C be on L.D., it is plain if lines parallel to AB and of equal 
length be drawn on G.P. within the P.P., they will appear 
shorter than AB. Find points M, N and join BM, CM, etc. 
Then AF = AB, GB = BC = AB, etc. Hence FG = AB, GH = 
BC, etc., also AF = BG, BG = CH, etc., and the squares AG, 
BH, CK, and DL will all be equal, and the only real right 
angles will be BCH and DCH ; all the other angles, FAB, 
ABG, etc., though really right angles will not appear so. 
(Fig. 14.) 



THE squab: 



29 




30 



DRAWING. 



To represent a square perpendicular to G.P. and P.P. 

Let H.L. and B.L. be drawn, also L.D., and find X, Take 




any points A, B, C on B.L. at equal distances, and at each 
erect a perpendicular equal to AB or CD ; join EH, which 



THE SQUARE. 31 

will .be parallel to AD; join A C.Y., B C.Y., C C.V. and 
J) C.Y. Take a point a at a distance from A equal to AB 
(on left), and join AX to cut A C.Y. in K ; draw KN parallel 
to AD, intersecting the vanishing lines in L, M, N ; at K, L, 
M, N, erect perpendiculars to meet the vanishing lines in Q, 
P, R, S ; join QS. Now, since AB = AE, AF is a square, 

(a) AF is said to be drawn to right, perpendicular to G.P., 
and touching or coincident with P.P. 

(b) Since A(X = AK = QE, then KE is a square drawn to 
right, perpendicular to G.P., perpendicular to and touching 
P.P. and parallel to L D. 

(c) Since AK = AB = KL = LB, AL is a square drawn to 
right, resting on G.P., perpendicular to and touching P.P. 

(d) Since AB = AK = QE = PF, QF is a square drawn to 
right, parallel to G.P. and raised above it, and perpendicular 
to and touching P.P. and parallel to L.D. 

The cube AP or BR will show the square in every position 
in parallel perspective, provided it touch the picture plane. 

(e) Similarly, square QL is drawn to right, perpendicular to 
G.P., parallel to P.P., and within it, (Fig. 15 ) 

Note. — If a figure be drawn parallel to the P.P. it will 
always be drawn in its true shape, but smaller, hence QL 
will be a true square. In above figure, take points a, 5, c, d 
etc., at equal distances to the left; erect perpendicular de at 
rf, and equal to do. Join e C.Y. and (i C.Y., also cX, 6X, 
aX, to meet d C.Y. in /, g, h. Through /, ^, h draw 
parallels to de^ meeting e C.Y. in I, m, n. Then, since de^ fl^ 
etc., are parallels between vanishing lines, they are all equal 
to each other and to dc^ ch^ etc., for de^dc. But df=^dc^ and 
fg = ch^ etc., hence e/J Ig^ mn are equal squares, and are said 
to be drawn perpendicular to G.P. and also to picture plane. 
One of them, e/J touches the P.P., the others are within it. 
It will be seen that as the squares recede they appear smaller. 



S2 



DRAWING. 



Example 1. — Draw a square, side i\ i' to R, 4' within, 
lying on G.P., _L to P.P. H = 6^ L.D. = 4', scale ^ = V. 

Construct figure as before. Take AC one inch (4') and CD 
one inch, take AB to left one inch, join C C.Y., D C.V., 
B MP, A MP, cutting C C.V. in F, E. Draw EK and EG, 
parallel to AC, and meeting D C. V, in K, G : then EK is 
square required, for EG = EK = CD, EC = CA, and EF = AB. 
(Fig. 16.) 

CV WP 




Fig. 16. 



Example 2, — Draw a square, side i' and 6' to left, _J_ to 
P.P. and 3' within it, and _|_ to G.P. and 1' above it. 
Other specifications as in last example. 

Construct figure as before. Take C \y (6') to left of A, 
take B V (4') to right of A, erect CE _j_ to AC at C and l^ 
{r^') in height, and mark off* D Y (!') above C; join E C.Y. 
D C.Y., C C.Y., A MP and B MP. Through E, G draw EL 
and GS parallel to CE, and meeting vanishing lines in L, S 
and H, K respectively : LK is the figure required. Eor 



THE SQUARE. 



83 



LS = IIK-FG = AB-:4^, SE: = LH=ED = 4 and HF^DO 

= 1', also FC = AC = 6'^. (Fig. 17.) 




Fig. 17. 



Uxamph 3, — ^Dra^ a square 4' side directly in front, paral- 
lel to G.P. and 3' above it, and _|_ to P.P. and 3' within itw 
Same specifications as before. 

Construct figure and find M as before. From A measure 
off l'^ both ways to B and C (BC = 4') ; erect perpendiculars 
BF and €G, each f (3'); join FG, B C.T,, F C.Y,, and G C.Y. 
From B measure off BD f*' (3'), and from D measure off DE 
V (4') ; join DM and EM, cutting B C. V. in H, K, Draw 
HL, KK parallel to BF, and meeting F C.Y. in L, N; through 
L, I^ draw LJ and J^P parallel to FG : then LP is the 
square required. For FL = BH = BD = 3^, and NP = LJ = FG 
= BC = 4', L]^ = HK = DE = 4', andLH = FB = 3'; alsoSL = 
SJ = XF = XG = AB = AC = 2'. (Fig. 18.) 



DRAWING. 



^ 


^ / 


UJ 

o \ 

^ \ 

<c 07 o 


'^ 


y/^ y 


/ 


CO 


/ / 


VJ 














^ 


\ 




/ 




-J 


\ 


^^-^ 


CQ / 


•«4 


/ 

/ / 

/ / 
/ / 

/ / 
/ / 
/ / 
/ / 

1 f 

'/ 


/ 
/ 

-J 

CQ 



bo 



THE SQUARE. 



35 



Example ^. — Draw a square 4' side, 4' to right, lying on 
G.P., _]_ to P.P. and 4' within it, and within this square 
place centrally a square whose sides are 2'. 




Fig. 19. 



Is"oTE. — Figures are said to be placed centrally when their 
centres coincide and their like sides are parallel to each other. 
Concentric circles are always placed centrally with respect to 
each other. 

Draw H.L., B.L., L.D., as before, and find M. Measure off 
aA, ae^ and AE, each one inch; bisect AE and ae in C and 
c, and bisect CE, CA, ce and ca in D, B, d, h respectively. 
Join A C.Y., B C.y., C C.Y., D C.Y., aM, 6M, cM, c^M, and 
eM, cutting A C.Y. in f, g, n, k, I; and through these latter 
points draw parallels to AE, cutting E C.Y. in t, s, r, q and J. 
U will form the outer square, and XX the inner square. 
For ac = BD = 2\ and kg=^dh^ 2\ and /A = Aa and ft = AE, 
etc. (Fig. 19.) 



36 DRAWING. 

Note. — If aM be joined, it will always pass through j if 
the figure be a square. It will also pass through X and X. 

Exercise IY. 

(H = 6', L.D. = 4', scale i" = r.) 

1. Draw a square 4' side, 3' to left, resting on G.P., _l_ to 
and touching P.P. 

2. Draw a square 5' side, directly in front, lying on G.P., 
I to and touching P.P. 

3. Draw a square 6' side, _J_ to G.P., _]_ to P.P., touching 
both, and 6' to left. 

4. Draw a square 3' side, parallel to G.P. and '2' above it, 
4' to R., perpendicular to P.P. and 2' within it. 

5. Draw a square 4' side, parallel to P.P. and i' within it, 
resting on ground plane, left corner touching L.D. 

6. Draw a square i' side, parallel to G.P. and 2' above it, 
right corner V to right; square to have side _l_ to P.P. and 
1' within it. 

7. Draw a square 4' side, J_ to G.P. and V above it, 3' to 
right, J_ to P.P. and 2' within it. 

8. Draw a square 6' side, directly in front, _|_ to P.P. and 
2' within it, parallel to G.P. and 4' above it; and place a 
square of one-fourth its area centrally within it. 

9. Draw a square 3' side, 4' to right, _J_ to P.P. and 3' 
within it, _l_ to G.P. and just its own height below it. 

10. Draw a cube (edge 4') touching P.P. 4' to left, resting 
on G.P. 

11. Draw a cube (4' edge) directly in front, and to st on 
G.P., one edge parallel to P.P. and 2' within it. 



TilE OBLONG. 



37 



The Oblong. 

An oblong is a figure whose opposite sides are equal and 
parallel, and whose angles are right angles. 

The drawing of the oblong diders little from that of the 
square, care being required only to distinguish the sides. 




Fig. 20. 



Example 1. — Draw an oblong 3' x 2^ lying on G.P., J to 
P.P., side 3', parallel to^ and 2' within it; oblong to be i' to 
right. Specifications as before. 



38 DRAWING. 

Construct figure as before. Find M, and measure off AC 4' 
= (distance to right), CD = 3^ (length of side), and measure off 
BE 2' (breadth); measure off BC=:2' (distance within P.P.). 
Join EM, BM, C C.Y., D C.V.; through H, F draw HK, FG 
parallel to CD, meeting D C.Y. in G, K : then FK will be 
oblong required. For FG = HK = CD = 3', GK = FH = EB = 
2', and HC = CB = 2'. (Fig. 20.) 

Exercise V. 

(H = 6', L.D. = 4', scale ^ = T.) 

1. Draw an oblong 4' x 2' _]_ to G.P., and 4' to left, J_ to 
P.P. and 3' within it ; oblong to rest on end. 

2. Draw an oblong 7' x 5', standing on end on G.P., parallel 
with P.P. and 2' within it, directly in front. 

3. Draw an oblong 3' 6' x 4' 6\ parallel to G.P. and 2' 6' 
above it, right corner 3' 6^' to left ; end of oblong to be paral- 
lel to P.P. and 6" within it. 

4. An oblong 6' x 4' is buried in the ground to a depth of 
2' ; it is parallel with LD and 4' to right, and 2' within P.P. 

5. An oblong 6' x 4', with side resting on G.P. 4' to left, 
and _j_ to it, intersects a square of 4' side, resting 6n G.P. 
and parallel to P.P. ; the oblong divides the square into two 
equal portions, and the square divides the oblong into parts of 
4:' and 2' respectively, the greater portion being nearest. The 
square is 4' within the P.P. 

6. Draw an oblong 4' x 2', end touching P.P. 4^ to R. ; 
oblong lying on G.P. 

7. Draw an oblong 5' x 3', lying on G.P. directly in front; 
end parallel with P.P. and 2' within it. 

8. A wall, whose height is 8', begins at a point 10' to left, 
and stretches inwards indefinitely ; at distances of 10' and 20' 
doors 5' X 3' are made in it. Draw it. 

9. Draw an oblong 6' x 4', lying on ground, sides parallel 
to P.P. and 2' from it, oblong 3' to right ; and within it place 
an oblong 4' x 2' centrally. 

10. Draw an oblong 3' x 2', lying on G P. 4' to left, end 
parallel to P.P. and 2' from it ; and about it draw an oblong 
5' X 4' centrally. 



THE TKIANGLE. 



^9 




Ttie Triangle. 

A triangle is a figure enclosed 
by three straight lines ; these 
may be of uniform length, but 
the length of any two taken 
together must be greater than 
the third. 

The triangle cannot be con- 
veniently drawn alone, but is 
drawn with reference to an ob- 
long. 

Triangles are equilateral, with 
three equal sides ; 

Isosceles, with two equal sides; 

Scalene, with unequal sides. 

Right angled, or containing a 
right angle ; 

Obtuse angled, or containing 
an obtuse angle ; 

Acute angled, or containing 
three acute angles. 

To draw a triangle we first 
draw a plan^ showing the tri- 
angle and its bounding oblong. 

Thus, 1 would be a plan for 
an equilateral triangle ABC, in 
which D A = AE = BF = FC. 

2 would be a plan for a right- 
angled triangle, in which the 
angle hoc is theright angle, with 
he as hypothenuse ; the segments 
5/, fc "of the hypothenuse would 
determine the position of a. 



40 



DBA WING, 



In 3, the triangle DAC being given, CE would be drawn 
perpendicular to DA, and CB parallel to DA ; also DB and 
AF would be drawn parallel to CE : then position o£ the 
angles at A and C would be easily dertermined. 

Example 1. — Draw an equilateral triangle, 2t' to b, side, 
lying on ground, one side touching P.P. near left angle, 4' to 
right. Height 6', distance 4\ scale ^ = 1'. 




fig. 22, 



Draw H.L., B.L. and L.D. as before. Take A i' to right of 
S, and B 3' to right of A; join A C.Y. and B C.Y.; on AB 
describe equilateral triangle ABE (below AB). Draw AD, 
BF at right angles to AB, and through E, draw DF parallel 
to AB, and meeting AD in D and BF in F ; draw EG paral- 
lel to AD. Join G C.Y., and with centre A and distance 
AD, describe arc DL, cutting B.L. in L. Join UM, cutting 



THE TRIANGLE. 41 

A C. V. in N ; through N draw NK parallel to AB, cutting 
G C.Y. in C and B C.Y. in K. Join CA and CB : then 
ABC is the triangle required. For AB = AE and AD = AL 
- AN = CG = height of triangle ; also NC = CK = AG = GB = 
DE = EF. Then AC = AE = EB = BC, for they are diagonals 
of equal oblongs. (Fig. 22.) 

In this figure the vertex C is directed away from the eye. 

If we joined NG and KG we would have a similar triangle, 
but with vertex G directed towards the observer. 

ExaTTiple 2. — Draw an isosceles triangle whose sides are 
2', 3' and 3' respectively, lying on ground plane, vertex 
directed towards the spectator, base parallel with P.P. and 5' 
away from it j near angle 2' to left. 

Draw B.L. and H.L. as before, find M. Take K 2' to left 
of X, and G 2' to left of K ; bisect GK in H j draw GR, HS 
and KT at right angles to B.L., and b' in length ; join RST. 
Take TP 3', and with centre T and distance TP describe arc 
PN, cutting HS in N : join KT and KR : KRT will be the 
plan of the triangle. 

Through N draw UNQ parallel to RT, cutting GR in U 
and KT in Q ; with centre K and distance KQ describe arc 
QL, cutting B.L. in L; and with same centre and distance 
KT describe arc TY, cutting B.L. in Y. Join G C.Y., 
H C.Y. and K C.Y. ; and join also LM and Y]\I to cut 
K C.Y. in E and C. Through E draw EAF, and through 
C draw CDB, each parallel to B.L., and cutting H C.Y. in A 
and D respectively ; join BA and AC : then ABC will be the 
triangle required. For CK = KY = KT = 5', and EC = LA" = 
QT = NS = height of triangle, and EK = KL==KQ. Hence 
AD = NS and AH = HN. But since triangle is isosceles, GH 
is made equal to HK; hence BD = DC = GH = HK, then CA 
= NT = NR = BA. (Fig. 23.) 




Example S, — Draw an equilateral triangle, each side 4', one 
side on ground plane parallel to L.D. ; triangle placed _|_ to 
ground plane and touching P.P. 3' to right. 

Draw H.L., B.L. and L.D., and find M. Take point D 3' 
to R., and B 3^ to right of D ; on DB describe the equilateral 
triangle DGB. Bisect DB in E, join EG ; through G draw 
FH parallel to B.L., and through D, B draw DF, BH parallel 
to EG, cutting FH in F, H. Join B C.V., draw EM, DM, 
cutting B C. Y. in N and A. With centre B and distance BH 
describe arc HK, cutting BK, a perpendicular on B.L., at K ; 



THE TRIANGLE. 



43 



join K C.Y. Through A and N draw parallels to BK, meet- 
ing K C.Y. in L and C ; join AC, CB. Then ABC is the 
triangle required. For LA = CIvT = KB = BH = EG = altitude 
of the triangle; and AN = NB = DE = EB, and AB = DB = 




length of side ; then C corresponds to G, and AC, CB to DG 
and GB respectively. (Fig. 24.) 

Example 4- — Draw a right angled-triangle whose hypothe 
nuse is 4' and one of the other sides 3' ; the right angle is 



44 



BRAWINa 



directed away from the observer, and is 5^ to the left, and 
the hypothenuse is 3' to the left. The triangle lies on the 
ground with the hypothenuse in contact with the B.L. 




Fig. 26. 



Draw H.L., B.L. and L.D. as before. Find M, and take a 
point D, 5' to left, 0, 3' to left, and B, 4' to left of C. On 
BC describe the semicircle BGC. (The angle BGC in a semi- 
circle is always a right angle.) Join DG, B C.V., T> C.V. and 
C C.V.] through G draw EF parallel to BC, and through B, C 
draw BE, OF perpendiculars on EF, at E and F. With 
centre C and distance OF, describe arc FL, cutting B L. in L; 
join LM, cutting C 0. V. in H ; through H draw KAH 
parallel to BC, cutting D C.V. in A ; join AB and AC : then 



THE TRIANGLE. 45 

ABC is the triangle required. For AB = BG and AC = CG, 
and AD = DG = CF = CL, and KA = BD and AH = DC ; then 
KH = BC, and A corresponds to G; then angle BAC cor- 
responds to angle BCG. (Fig. 25.) 

Exercise YI. 

(H=6', LD = 4^ scale i'^==r.) 

1. Draw an equilateral triangle, 3' side, lying on G.P., one 
side parallel to L.D., vertex directed to the left and distant 
from the L.D. 4' ; the triangle touches P.P. 

2. Draw an equilateral triangle parallel to P.P. and 3' from 
it, _J_ to G.P., near angle 2' to left, triangle 3' to side, one 
side on ground. 

3. Draw a triangle whose sides are 4', 5', 6^, respectively, 
6' side on ground, _j_ to P.P. and 2' within it ; triangle 3' to 
right. 

4. Draw an isosceles triangle whose base is 3^ and each 
equal side 4', lying on ground plane directly in front, vertex 
directed towards P.P. and 1' from it. 

5. Draw a right-angled triangle whose hypothenuse is 5', 
and the perpendicular on it from the right angle divides it 
into segments of 3' and 2'. The triangle is parallel with G.P. 
and 4:' above it, and the vertex is directed away from P.P. at 
a distance of 4^ The right angle is directly in front, and the 
larger segment is to the right. 

6. Draw a triangle, each side being 3', _|_ to G.P. and 1' 
above it, with a side parallel to it. The triangle is 4' to left, 

I to P.P. and 2' within it. 

7. Draw an isosceles right-angled triangle, the equal sides 
being 3'; one equal. side is parallel to G.P. and 2' above it. 
The triangle is 2' to right, parallel to P.P., and 2^ within it. 

8. Draw an isosceles triangle, base 4', equal sides 3' each, 
directly in front, paxallel to P.P. and 4' from it ; vertex touches 
ground, and base is parallel to it. 

9. An equilateral triangle, each of whose sides is 4', lies on 
the ground, vertex directed away ; one side parallel to P.P. 



46 DRAWING. 

and 2' from it. Triangle 2' to right. Within this place cen- 
trally a similar triangle whose sides are 2'. 

10. An equilateral triangle, each of whose sides is 4', is _j_ 
to ground and also to P.P., which it touches at a point 5' to 
left; the vertex of the triangle is directed downwards, and 
one side is horizontal. Tlie triangle is buried one-fourth in 
the ground. Draw it. 



THE HEXAGON. 



47 



Ttie Hexagon. 

A hexagon is a figure bounded by six sides. When the 
sides and angles are all equal the hexagon is regular. Only 
regular hexagons will be considered here. 

In drawing a hexagon, a plan is made somewhat similar to 
that for the triangle. The following is the general form of 
the plan. 




Fig. 26. 

Let B.L. =base line, and AB given side of hexagon, and in 
given position. Bisect AB in C, make AD and BE each equal 
BC ; on DE describe equilateral triangle DFE, and construct 
oblong DL. Bisect DK in G ; draw GH parallel to KL ; join 
AG, GM, BH and HN. With centre D and distances DG 
and DK describe arcs to cut B.L. in X and P respectively. 
Like triangles, hexagons may lie flat on the ground, with a 
side parallel to the P.P. or to L.D., or they may be drawn 
perpendicular to the ground plane, with a side or an angle 
touching it ; and a side or an angle may touch the P.P. or be 
within it. (Fig. 26.) 

If a side touch the P.P., the extremity of it determines the 



48 



DRAWING. 



distance of the hexagon to the right or left. When an angle 
touches the P.P., that point determines the distance ; so also 
when the figure is within the P.P. 

Example 1, — Draw a hexagon, each side 2^, lying on G.P., 
one side coincident with P.P. and 4^ to R;. II = 6', L.D. = 4', 
scale ^'' = r. 




Let H.L., B.L. and L.D. be drawn. Take C.Y., find O and 
MP; take AB 4' to E. of Q, make AB = 2^ On AB con- 
struct equilateral triangle ABH, and produce to Y, Z, making 
HZ and HY each equal BH ; join AY and YZ, and produce 
YZ both ways. Through H draw GHW parallel to YZ ; 
through C draw CE parallel to AY ; draw also DWF parallel 
to AY. Join C C.Y., A C.Y., B C.Y. and D C.Y. ; also 
L MP and K MP, cuttin^ C C.Y. in N and P. Through 
N, P draw NT and PY : parallel to AB ; join SP, PA, TY 
and VB, then STYBAP is the hexagon required. For PC = 



THE HEXAGON. 



49 



CK = CG, and NP = LK = GE. Then SP = GY, also PA = 
AG = GY; similarly, TY = YB = B\y = Y/Z = AB, and QA 
= 4'and AB = 2^ (Fig. 27.) 
Example 2, — Draw a hexagon same as in Example 1, but 
I to G.P., and with an angle touching P.P. 4' to L. 




Fig. 28. 



Here take E 4' to left, and for plan of the hexagon we 
may proceed as follows : — Bisect EQ (4') in Z, also EZ, QZ 
in A, B respectively ; with centre A and distance AB 
describe arc BC ; similarly, draw arc AD. Draw EC and 
QD, perpendiculars to AB at E and Q respectively, meeting 
arcs in C and D ; produce EG to F, making CF = EC. 
Erect perpendicular EN and make it equal EF, and make 



50 



DRAWING. 



ES = EC; join N C.Y., S C.Y. and E C.Y., also AMP, 
B MP and Q MP, cutting E C.Y. in K, L, M. Through 
K, L, M draw paraUels to EN, cutting IST C.Y. in O, P, R ; 
join OS, SK, LT and TP : then OPTLKS will be hexagon 
required. For ES === EC and EN = EF, also KL - AB and EM 
= EQ. Then OS = SK = AB ; and PT = TL = BD. (Fig. 28.) 
Example S, — Draw a hexagon, 2' side, resting on G.P. _[_ 
to P.P., having one side coincident with it and 4' to right. 
H = 6', L.D. = 4; scale ^ = V. 




Fig. 29. 



We proceed as follows : — After drawing H.L. and B.L. and 
finding positions of O, MP and C. Y., take A 4' to right of S ; 
at A erect perpendicular AC = 4', and measure off AE = P and 
DE = 2'. Produce CA to B, making AB = AC and FX = DE ; 
make FG = FX and GH = GA ; join HKMP, GLMP, A C. Y., 



THE HEXAGON. 



51 



jD C.Y., E C.Y. and C C.Y. At K, L erect perpendiculars to 
meet vanishing lines in N and M ; join MR, MD, LE and 
LP : then RMDELPR will be hexagon required. For AK = 
AH and AL = AG = FX ^ 2', and DE = RP and ML = ISTK = 
CA. Then RM = MD - DE = 2', etc. (Fig. 29.) 

Example ^. — Draw a hexagon, 2' side, lying on ground, 
near side parallel with P.P. and 2' within it; hexagon to be 
directly in front. H = 6', L.D. = 4^ scale \;' = T. 




52 DRAWING. 

Draw H.L., B,L., and find O C.V. and MP as before. 
Make SY and ST each = 1', and make TX and YA each = T ; 
make Ar = 2^, and draw it _]_ to AX; draw XP, TY, YZ 
parallel to AF. Make RG^RL or TY, and GZ also = RL, 
and complete the oblong MH. Make Aa = AF, Ab = AG and 
AE = AH; join EMP, hMV, aMP, and A C.Y., Y C.Y., 
T C.Y. and X C.Y. Through intersecting points D, C, B, draw 
parallels to AX, meeting X C.Y. in e, /, g, respectively ; join 
cf, /k, dC and Cm, completing the hexagon : then AB = Aa = 
T, and BC = CD = a6 = 6E = GH, and hence Cm = GZ and dC 
= YG = 2^, etc. (Fig. 30.) 

Exercise YII. 

In these examples take PI = 6', L.D. = 4', and scale ;|" = 1' ; 
but a scale of ^'^ = V may be used if thought more convenient. 

1. Draw a hexagon, side 2', lying on ground plane, one side 
perpendicular to P.P., and an angle touching it at a point 4' 
to right. 

2. Draw a hexagon, side 3', standing on edge, _J_ to ground 
plane and P.P., and having an angle touch the P.P. 3^ to left. 

3. Draw a hexagon, side 2', standing on edge, parallel to 
P.P. _l_ to ground plane, directly in front, and 3' away. 

4. Draw a hexagon, 2' side, lying on G.P., one side parallel 
to P.P. and 3' away ; hexagon 4' to left. 

5. Draw a hexagon, 3' side, resting on an angle _|_ to 
G.P. and P.P., one side parallel to P.P. 4' to right and 4' 
within it. 

6. Draw a hexagon whose edge is coincident with that of a 
square, and lying in same plane. The square is 2' to the side, 
and is placed _j_ to P.P. and G.P. 2' to right and 2' within. 

7. Draw a hexagon, 3^ side, parallel to G.P. and 4' above 
it, 4:' to right, 3' within P.P., and one side parallel to P.P. 

8. Draw a hexagon 3' to side, 4' to right, one angle touch- 
ing P.P.; hexagon to be __[_ to P.P. and parallel to G.P. 



THE KLXAGON. 63 

9. Draw a hexagon 3' to a side, directly in front, lying on 
ground plane, one angle touching P.P., and sides _]_ to it. 

10. Draw a hexagon about an equilateral triangle lying on 
G.P., vertex directed away from observer ; the triangle is 3' 
to a side, and one side is parallel to P.P. and 3' within it. 
The vertex of the triangle is 4' to left. 

11. Draw a hexagon, 2' side, placed _[_ to P.P. and G.P., 
4' to right, lower side parallel to ground plane and 4' above 
it. 

12. Draw a hexagon, 4' side, lying on Gr.P., near side paral- 
lel to P.P. and 2' within it; hexagon to be 4' to right. 
Within this draw (centrally) another hexagon, whose sides 
shall be 2' in length. 

13. Draw a hexagon, side 2', parallel to P.P. and 2' within 
it, lying on G.P. directly in front. 

14. Represent a hexagon, side 3', half buried vertically in 
the ground, one side parallel to G.P.; hexagon __|_ to P.P. 
and 3' within it, 4^ to left. 



54 



DRAWINv 



The Octagon. 

The drawing of an octagon diflfers but little from that of 
the hexagon, we shall, therefore, merely show the plan. The 
following are methods of drawing the plan: — 





C A 


OS D 




H 


v% 


\ 


v^ 


P 


K 


<) 




N 




i L 




Af^ F 





Fig. 31. 



Let AB = given side (on B.L.) ; bisect AB in O, and draw 
OG _J_ to OA and equal it, and describe semicircle AGB; 



THE OCTAGON. 



OD 



join AG and GB ; with centres A, B, and distances equal to 
AG, describe arcs to cut AB produced in C and D ; on CD 
describe square CDFE. With centre A and distance AB 
describe arc to cut CE in H ; find P similarly ; with centres 
H and P and distances equal to PIA, describe arcs to cut CE 
and DF in K and N respectively, and with same distances 
describe arcs to cut EF in L and M ; join AH, KL, MN and 
BP, which will complete the octagon, (Fig. 31.) 
Another way : 

a 




56 DRAWING. 

Let AB = given side ; take centre O, and with distance AO 
describe circle AGD ; draw diameter CD _J_ to AB. Join 
CA and produce it. With centre A and distance AB describe 
arc to cut CA produced in E; then draw EF _J_ to BA pro- 
duced, and produce FE, making EG = EA, etc. (Fig. 32.) 

Note. — In parallel perspective a hexagon or an octagon 
must be supposed to have a side parallel or perpendicular to, 
the picture plane. 



THE CIKCLE. 57 



The Circle. 

Hitherto we have been dealing exclusively with straight 
lines, in so far as the appearance of figures is concerned ; we 
now proceed to represent curved lines in perspective. It is 
evident that a curve cannot be correctly represented, without 
the aid of straight lines. 

There is only one -position in which a circle will appear true 
to the eye, and that is, when the eye is in a line exactly per- 
pendicular to its plane, at its centre. In all other positions it 
will appear an ellipse, varying from a circle to a line. If, for 
instance, we place a hoop on the ground, and look at it 
directly, it will appear true, but if turned on an imaginary 
axis it will assume the form of an ellipse. The height or 
diameter of the hoop corresponding to the imaginary axis will 
remain the same, while the diameter at right angles to it, or 
the revolving axis, will diminish, till at length it is a mere 
point. Hence, to know the appearance of a circle not viewed 
directly, we must know the angle the eye makes with its 
plane, or its appearance in relation to some figure easy of 
representation, contained by straight lines. 

Now, a square answers admirably for this purpose, for if we 
draw the diameters of a square, and then draw a circle so as 
to touch its sides at the extremities of the diameters, we can 
without much difficulty represent the circle, for we will have 
four j^oints as guides. If, however, the diagonals also, of the 
square be drawn, the four points where they cut the circum- 
ference of the circle will furnish additional points, so that we 
will have altogether, eight points _f or guidance in drawing the 
circle. Thus, — 
5 



58 



DRAWING. 




Let KLMO = given square, draw diameters and diagonals, 
and inscribe circle cutting the diagonals in A, C, G, E; 
join AC and GE, and produce them to meet KL in S, S ; join 



THE CIRCLE. 



59 



K C.Y., S C.Y.,. etc.; also mafe KIl = KB and RX^BO or 

KB. Join X MP and B MP, and where they meet K C.Y. 
draw parallels to KL : then hf will represent the diameter 
BF, and it will be easily seen that a, 5, c, d^ e^f^ g^ h will cor- 
respond with A, B, C, D, E, F, G, H respectively, and the 
curve traced between them will represent the circle (in this 
case) lying on the ground plane. (Pig. 33.) 

Lxample 1. — Draw a circle, diameter 4', lying on G.P., cen- 
tre 4' to right and 2' within P.P. H = 6', L.D. = 4', scale 
X" _ ] / 

c v_ i^LP 




Fij?. 34. 



Here ED = 4^, EA = 2\ DB = 2'. Describe semicircle ASD, 
draw AG and BM J_ to AB, and draw GSM through S, 



60 



DRAWING. 



parallel to AB ; make AE = AG and AF = AB; join F MP 
and E MP, also A C.Y., D C.Y., B O.Y. Join DG and DM, 
cutting curve in H and K. Draw HO and KL parallel to 
AG; join C C.Y. and L C.Y.; complete square NABP, draw 
diagonals; then on the eight points thus shown draw the curve 
required. (Pig. 34.) 

Example 2. — Draw a circle touching P.P. M to left, stand- 
ing on G.P. and _[_ to it and P.P. ; circle to be M in diameter. 
H = 6', L.D. = 4^ scale ^'' = 1^ 




Fig. 85. 



Here take A 4' to left, bisect it in G ; describe semi- 
circle, and complpt^ oblong APQB ; join GP, GQ, cutting 
curve in P, S ; dviw RF, SH _|_ to AB ; join A C.Y. Erect 
at A the peri^-^ndicular AC = AB, and bisect it in X; join 
C C. Y., X r. 7., also F MP, G MP, H MP, B MP. At points 
of section v, L, M, E draw parallels to CA; draw diagonals 



THE CIRCLE. 



61 



CE, AD, and trace curve between the eight marked points. 
(Fig. 35.) 

Example S. — Draw a circle, diameter 4^, directly in front, 
lying on G.P., centre i' within P.P. H = 6^ L.D. = 4', scale 




62 



DRAWING. 



Here take OA, OB, each = 2'; upon AB describe square 
ACDB; bisect BD in E, and draw EM parallel to CD; 
describe semicircle and find points F, G, as already shown. 
Make BK = BD, BH = BE and KL = DE; join L MP, 
K MP, etc.; also A C.Y., G C.Y., etc.; and on eight points 
thus formed describe circle required. For BX = BH = BE 
= 2', and XZ = HL = BD = 4', etc. 

Example ^. — Draw a circle, diameter = 4', _J_ to G.P., 
parallel with P.P. and 2' within it; centre of circle b' to 
right. H = 6', L.D. = 4', scale ^ = !'• 



CV_ 



MP 




Here take A 5' to right of O ; erect the perpendicular AB. 
2' = radius of given circle; join B C.Y. and A C.Y.; take 
2' to left of A; join MP, cutting A C.Y. in D. Draw DE 
parallel to AB ; then with centre E and distance ED describe 
circle required. For AD = AC = BE; then E is 2' within 
P.P. and 5' to right, also DE = AB = 2', etc. (Fig. 37.) 



THE CIRCLE. 63 

Exercise YIIL 

(H = 6^ L.D. = 4'; scale ^ = V.) 

1. Draw a circle, diameter 4^, resting on ground plane and 
touching P.P. at a point 4' to right. 

2. Draw a circle, diameter 4', resting on G.P., centre 4' to 
left and 4' within P.P. 

3. Draw a circle, diameter 4' its plane perpendicular to 
G.P. and touching it at a point 4' to left; the circle is per- 
pendicular to P.P. and touches it. 

4. Draw a circle, diameter 4^, lying on ground plane directly 
in front, centre 5' within P.P. 

5. Draw a circle, diameter 4^, parallel to G.P. and 2' above 
it, placed with centre 4' to right and 4' within P.P. 

6. Draw a circle, diameter 4', coincident with P.P. and 
touching G.P., centre 4' to right. 

7. Draw a circle, diameter 4^, plane perpendicular to G.P. 
and P.P.; the centre of the circle is 5' to left and 3' within 
P.P. 

8. Draw a circle, diameter 4', parallel to G.P. and 9^ above 
it, directly in front, centre 4^ within P.P. 

9. Draw a circle, diameter 4', parallel to P.P. and 6' within 
it; centre of circle V to right and 3' below" G.P. 

10. Draw a circle, diameter 6', lying on G.P., centre 4' to 
right and 4^ within P.P., and wdthin it draw a concentric 
circle of 3' diameter. 

11. Draw a quadrant, radius 2', lying on G.P., vertex 
directed away, and placed 4' to left and 4' within P.P.; the 
radii make an angle of 45° with P.P. 

12. Draw a circle, diameter 4', buried vertically in the 
ground to a depth of P; the circle is perpendicular to P.P., 
and its centre is placed at a point 5' to left and 3' within 
P.P. 



64 DRAWING. 



SOLIDS 

It is expected that the pupil will have drawn all the figures 
mentioned in the exercises. Unless the problems have been 
thoroughly understood, comparatively little progress can be 
made in the perspective of solids. 

Solids may be classified thus : 

I. Those contained by plane surfaces. 

II. Those partially or wholly contained by convex surfaces. 
They are sometimes classified as solids with Developable, or 

with Undevelopable surfaces. 

Those belonging to Class I. are Cubes, Plinths, Parallelo- 
pipeds. Prisms, Pyramids, Wedges, and Frusta. 

Of those contained by convex surfaces in part, are Cones, 
Cylinders, Hemispheres, and frusta of Cones. 

Those contained wholly by convex surfaces are Spheres, 
Spheroids, Ellipsoids, Cylindroids, Spindles, etc. 

All the latter have undevelopable surfaces, ^.6., they cannot 
be straightened out to a plane surface. 

Solids contained by plane surfaces may be subdivided into : 

1. Those rectangular throughout. 

2. ir partly rectangular. 

3. ff wholly oblique. 

The latter class of solids cannot be readily drawn in per- 
spective, and will not be treated of, here. 

Of (1) are Cubes and Plinths or Parallelopipeds. 

A cube is a solid contained by six equal squares, and aU 
its angles are right angles. 

A plinth is a solid contained by three pairs of equal and 
similar oblongs. Each pair of surfaces may be equal or 



SOLIDS. 65 

unequal to one or both of the other pairs, but the angles are 
right angles. 

(2) A prism is a solid contained by two regular polygons 
whose planes are parallel to each other, and whose like sides 
are joined by rectangular planes, 

A pyramid is a solid formed by joining the angles of a 
triangle, square, etc, with some external point. If the exter- 
nal point be vertically above the centre of the pyramid, the 
pyramid is said to be right ; if in any other position, oblique, 

A frustum of a pyramid is the part remaining after a 
smaller pyramid is cut off loj a plane parallel to the base. 

Of solids with convex surfaces : — 

A sphere is formed by the revolution of a semicircle around 
the diameter, which remains fixed. 

A cone is formed by the revolution of a right-angled 
triangle around one of the containing sides, which remains 
fixed. 

A cylinder is formed by the revolution of an oblong around 
one side, which remains fixed. 

A spheroid is formed by the revolution of a semi-ellipse 
around one of the axes, which remains fixed. 

If the fixed axis be major^ the spheroid is prolate • if 
^QfiinoT^ oblate. 



DRAWING. 



Tine Cube. 

The drawing of the cube is so simple that a single example 
will suffice for its explanation. 




THE CUBE. 67 

Example 1, — Draw a cube, edge 4', placed on G.P., one side 
parallel to P.P. and 2' within it, near angle 3' to right. 

H = 6; L.D. = i\ scale ^' = V, 

Draw B.L., H.L., and O, C.Y. and find IMP as before. 
Take A 3^ to right, and B 4' to right of A ; take 2' to left 
of A, and D M to left of ; join A C. Y., D MP and 
C MP, cutting A C.Y. in F and E. On AB describe square 
ASMB; join S C.Y., M C.Y. and B C.Y.; through E and F 
draw parallels to AS, meeting S C.Y. in H and G ; through 
G, H draw parallels to SM, meeting M C.Y. in K and L. 
Draw LN from L J_ to MB, and EN" from E parallel to AI5. 
This will complete the required cube. For XA = 3', and 
AE = AC = 2', AF = AD and EF = CD = AB = 4'; also FG== 
EH = AS = 4', and GK = HL = EN = AB = 4': thenEF = GH 
= KL=4', etc. (Fig. 38.) 



68 



DRAWING. 



The Pllntti. 




The plinth differs from 
a cube only in the rela- 
tion of its dimensions; 
the principle employed 
in drawing them is the 
same, but a particular 
side of the plinth is men- 
tioned in reference to 
the P.P. or G.P. In the 
cube this is quite un- 
necessary, as all the sides 
are equal. 
oi Example 1, — Draw a 
.fcjD plinth whose dimensions 
^ are 4' x 3^ x 2' (4' long, 
W wide and 2' thick), 
the side 4' x 3' rests on 
the ground plane, and 
side 3' X 2^ is parallel to 
P.P. and 2' from it; the 
plinth is 2' to the left. 
H = 6^L.D. = 4',scale 

Draw H.L., B.L. and 
find C.Y.,OandMPas 
before. Take B, 2' to 
left, and A, 3' to left 
of B; also M, 4' to 
right of O. On AB 



THE PLINTH. 



69 



construct the oblong ABDC, 3^x2'; join C O.V., D C.V., 
A C.V., B C.V.; also O MP and M MP. Prom points K, H, 
where B C. V. intersects O MP and M MP, erect KN and HG 
parallel to BD ; and through N, G draw NE and GP paral- 
lel to CD, Draw EL J_ '^<^ EN and KL J_ to EL, which 
will complete the plinth. For BK =^ BO and KH ^ OM ; then 
EP-NG-KH-OM-4>ndPG-EN-LK-AB-3^GH 
-NK-DB-2', etc. 

Example 2, — Draw a flight of four steps, each step 4' x 1' 
X r. The ends of the steps are coincident with the P.P., and 
i' to right. H - 6', L.D. - i', scale ^ ^ ^'* 




Fig. 40. 



Here take A, i' to right of E, and B 4^ to right of A ; on 
AB describe square ABDC, and divide it into sixteen equal 
squares ; join each angle, as shown in figure, with the C.Y. ; 
join also E MP, cutting A C. V. in P ; draw FG parallel to 
AC, GS J_ to GF, etc. Then FG - AK = 1 ', tmd GS = KJST 
= r, etc.; and AF = AE = AB = 4'. (Fig. 40.) 



70 



DRAWING. 



Example S, — Draw same, with ends perpendicular to P.P., 
one step being coincident with it and i' to left. 




Fig. 41. 

Draw H.L., B.L., and find C.Y. and MP as before. Take 
A, i' to left, and B, 4' to left of A ; on AB construct square 
ABCD ; join O MP ; and on AE, complete square ADFE ; 
then draw steps similar to preceding example. 

Exercise IX. 

(H = 6^ L.D. = 4; scale l'' = r.) 

1. Draw a cube, edge 4', 2' within P.P. parallel to it, and 
4' to right. 

2. Draw a cube, edge 3', directly in front, at a distance of 
3' from P.P., resting on G.P. 

3. Draw a cube, edge 5', parallel to G.P. and 2' above it; 
cube 3' to right, parallel to P.P. and touching it. 

4. Place two cubes, each 4' edge, on a line parallel to L D., 
cubes to be 4' apart, and nearest 4' from P.P. and 4' to left. 

5. Draw a cube, edge 4', touching P.P. and 4' to right; and 
place a cube, 2' edge, centrally upon it. 



THE PLINTH. 71 

6. Draw a plinth 6' x 4' x 2', side 4' x 2^ on ground, side 
6' X 2' parallel with P.P. and 4' to left ; figure to be 2' within 
P.P. 

7. Draw a slab 4' x 2' x 2', lying on ground directly in 
front, side 4' x 2' parallel to P.P. and 4' from it. 

8. Draw a slab 5' x 5' x 1' lying flat on G.P., side 6' x 1' 
parallel to P.P. and 3' from it; slab to be 4' to left. Place 
centrally on this slab a cube whose edge is 3'. 

9. Draw a cube, 2' edge, on each side of L.D., 2' from it, 
and touching P.P.; on these cubes place a slab 6' x 2' x 1' 
coincident with the cubes. 

10. A wall 8' high and 2' thick starts from a point on the 
P.P. 4' to the left, and runs straight forward to the horizon ; 
at distances of 6^ and 12' doors 5' x 3' are placed. 

11. Draw a cross whose beams are 7' x 1' x 1' and 5' x 1' x V 
respectively ; the cross-beam is placed at a height of 3^ The 
cross stands erect, its cross-beam parallel to P.P. and 4' within 
it ; the foot of the cross is 4' to right. 

12. Draw same, with end of cross-beam coincident with 
P.P., 4' to left. 

13. Draw same, lying on G.P., cross-beam _[_ to P.P. and 
its end coincident with it, 3' to right. 

14. Draw same, lying on ground directly in front, cross- 
beam directed away, end of main beam coincident with P.P. 

15. A circular table 4' in circumference is supported by 
four legs 2' high, which proceed from the edge of the table ; 
the legs form a square whose side is parallel to P.P. and 3' 
within it. The centre of the table is 4' to right. Thickness 
of neither table nor legs, taken into account. 

16. Draw a set of four steps, each 4' x 1' x 1', ends parallel 
to P.P. and 2' within it ; to be 5' to left, facing toward right. 

17. Draw same, ends perpendicular to P.P. and 2' within 
it, and 3' to right. 

18. Draw same, with steps descending as they recede ; back 
coincident with P.P. and 2' to left. 

19. Draw same, directly in front, steps ascending as they 
recede, and 2' within P.P. 



72 



DRAWma. 



TThie Prism, 

Prisms are square, triangular, hexagonal, etc., according to 
their ends or bases. 

The square prism may oe considerea as a mere modification 
of the plinth. 

To draw a prism, we have only to draw the two surfaces 
forming its ends, and join similar angles. 

Example 1, — Draw a triangular (equilateral) prism, length 
6', side of base 2', lying on G.P., one end perpendicular to 
P.P. and 3' to left; prism to touch P.P. II = 6^ L.D. = 4', 
scale i'' = r. 

MP CV 




Take B, 1' to left, A, 3' to left, and D, 6' to left of A. On 
AB describe equilateral triangle ABC; draw CL | to A3 



THE PRISM. 



73 



and bisecting it. Draw AE _[_ to AB and equal to CL; simi- 
larly draw DF; join F C.V., E C.V. and A C.V., also B MP 
and L MP ; through M draw MG- parallel to AE, meeting 
E C.Y. in G; draw GH parallel to EF; join HD, AG and 
GK, completing the prism. Then AK = AB - AC = BC == 2^ 
Hence AG - KG = AK, and GM - E A = LC - required height, 
and HG = FE = DA = 6', etc. (Fig. 42.) 

Example 2, — Draw a hexagonal pri^m (edge of base 2')' 
whose length shall be 4', one side touching P.P. 4^ to right; 
prism to stand on end. H = 6'; L.D. = 4'; scale, :|'' = 1. 




Fig. 43. 



74 DRAWING. 

Draw the plan in proper situation, as already explained; 
then draw the hexagons, one on G.P., the other 4' above it; 
then join similar angles in each, forming the required hexagon. 
(Fig. 43.) 



THE CYLINDER. 



75 



TThe C^rlinder. 

The drawing of the cylinder differs from that of the prism, 
only in the plan. Draw the circles, forming the ends, in the 
proper positions, and then draw tangents to them, forming 
the cylinder. 

Example 1, — Draw a cylinder, length M and diameter 4', 
lying on ground plane parallel to P.P. and 2' within it; the 
end of cylinder ^ to left* 




Fig. 44. 



76 



DRAWING. 



Example 2. — Draw a cylinder lying on G.P., 4' to right, 
having end perpendicular to P.P., and touching it; cylinder 
8' long and 3' in diameter. 




Ta^^e A, 4' to right and E, i' to left; make AB = 1^-', and 
describe circle; join B C.Y. and A C.V., also E MP, and from 
C, draw CD parallel to AB, and describe smaller circle; then 
draw common tangents, completing the cylinder. (Pig. 45.) 



THE CYLINDER. 



77 



Example S. — The figure shows how to draw a common pail, 
showing staves and hoops. 




78 



DRAWING. 



Ttie Pyramid. 

We now come to consider solids, which are not wholly rec- 
tangular; they are cones and pyramids and their frusta. 

In speaking of the height of a pyramid or cone, we mean 
the distance from the vertex perpendicularly to the base, 
yhis is important, especially in frusta, where the slant height 
might be mistaken for the real height of the solid. 

Example 1. — Draw a pyramid, 8' high with square base, 
each side of which is 4', and touches P.P. 4' to right. H = 6', 
L.D. = 4^ scale 1'' = ; 

K L M 




Fig. 47. 



Take A, 4' to right and B, 4' to right of A; complete the 
square ABDC; draw diagonals intersecting in O; join O with 



THE PYRAMID. 79 

C.Y. and produce it backward to meet base line in E; at E 
erect perpendicular, 8' in height to M; join M C.Y.; through 
O, draw OX, parallel to EM; join XA, XB, XC, completing 
the pyramid. ^Now, OX = EM = 8', and this represents the 
vertical height. (Fig. 47.) 

It is not absolutely necessary to join O with C.Y. We 
may draw it to any point on the H.L., as K or S., and pro- 
duce it backward to F or A, and erect a perpendicular from 
either of these points; but it must be carefully remembered, 
that the so-found point K or L, must be joined to S or K 
respectively. Such lines, KS, LN, M C.Y., etc., will all pass 
through same point X, which may be considered as a locus 
for all such lines. For convenience, however, the line O C.Y. 
should be used, unless the solid be directly in front. 



80 



DRAWING. 



TTbie Cone. 

The drawing of the cone does not differ materially from 
that of the pyramid. The circle forming the base being 
drawn, and the position of the vertex found, it is only neces- 




THE CONE. 81 

sary to draw the tangents from it to the circle. We give a 
particular example: — 

Draw a cone whose base = 6' in diameter and slant height 6\ 
The cone lies on its side; pla^ie of base, _|_ to P.P., and the 
line joining the centre of base with the vertex is parallel to 
the "P.P. and 4' from it. The cone is 4' to the right. H = 6', 
L.D. = 4', scale l'' = r. 

Take A, 4' to right and C, 6' to right of A; on AC describe 
equilateral triangle ADC; draw DB perpendicular to AC; 
join A C.V., B C.Y., C C.Y.; take Y and E, 3' from O; join 
E MP, O MP and Y MP; through F, G, H draw parallels to 
AB; through S, R, K draw parallels' to DB; join PF, ISTG, 
MH; then in square PH describe circle; produce GB to L, 
and from L draw tangents LX, LX to circle, completing the 
cone. (Fig. 48.) 

Exercise X. — On the Prism and the Cylinder 

(H = 6', L.D. = 4^ scale Y = V.) 

1. Draw a prism 6' in length, triangular base, each side of 
which is 2'. The prism stands on end 4' to right, with one 
side coincident with P.P. 

2. Draw same, 3' to left, one side perpendicular to P.P. 

3. Draw same, directly in front, one side parallel to P.P. 
and 2' from it, vertex away. 

4. Draw same, lying on ground plane, perpendicular to P.P., 
3' to right and 3' within P.P. 

5. Draw a hexagonal prism 6' high, each side of base 2\ 
standing on end directly in front, one side touching P.P. 

6. Draw same, lying on ground parallel to P.P. and 2' within 
it; one end projecting 2' to right, and opposite end 4' to left. 

7. Draw same, lying on ground perpendicular to P.P., 3' to 
right and 3' within P.P. 

8. Draw a cylinder, diameter of base 4', height 6', lying on 
ground, parallel to P.P. and 4' within it; left end just in line 
with L.D. 



82 DRAWING. 

9. Draw a cylindrical vessel 4' feet in height and 4' in 
diameter, standing on end, touching P.P. 4' to left] show 4 
hoops at distances of 1' from each other, 

10. Draw a hollow cylinder, 4' in length, outer diameter 4', 
inner diameter 3', lying on ground perpendicular to P.P., 2' to 
riffht and touching: P.P. • 



Exercise XI. — On the Pyramid and Cone, 
(H = 6', L.D. = 4^ scale Y = 1'.) 

1. Draw a square pyramid 5' high, each side of base 3', 
standing on ground, directly in front, touching P.P. 

2. Draw same, 4' to right, parallel to P.P. and 4' within it. 

3. Draw same, standing on a 3' cube, parallel to P.P. and 
2' within it, M to left. 

4. Draw same, 3' above ground plane and parallel with it, 
4' to right and touching P.P. 

5. Draw same, with vertex downwards, base parallel with 
ground and P.P., vertex, 4' to left and 3' within P.P. 

6. Draw a cone, height 5', diameter of base 4', standing on 
ground M to right and 4' within P.P. 

7. Draw same, touching P.P., 3' to left. 

8. Draw same, directly in front, 3' from P.P. 

9. Draw same, standing on a cylinder 4' in diameter and 4' 
high, 3' to right and touching P.P. 

10. Draw same, placed centrally on a cylinder of 5' in diam- 
eter, directly in front and touching P.P. 

11. Draw a cone, base 4', slant height 4', lying on side; 
base perpendicular to P.P. and touching it, 4' to left; vertex 
directed toward left. 



FRUSTA. 



83 



Kru^ta. 

The dimensions of a frustum may be given by stating dimen- 
sions of each end, and vertical height. 




84 



DRAWING. 



Example 1. — 5)raw the frustum of a square pyramid, whose 
bases are 6^ and 4' square, respectively ; the frustum touches 
P.P. 4^ to right, height 4'. (Fig. 49.) 

Prom the above the method of drawing may be easily 
understood. 

Example 2. — A pyranaid with square base, each side of 
which is 4', stands on the ground plane 4^ to the left, touch- 
ing P.P. The pyramid is 8' in height; 3^ from the vertex 
the pyramid passes through a square plinth 4' x 4' x^l' placed 
parallel to the ground plane. The pyramid cuts the plinth 
centrally. H = 6^ L.D. = 4^, scale \;' = V. (Fig. 50.) 




Fig. 50. 



By a careful observation of the lines drav/n above, the 
method may be easily seen. 



FRUSTA. 85 

Exercise XII. 

(H. = 6^ L.D. = 4; scale Y = !'•) 

1. A pyramidal frustum with square base, and vertical 
height 4', touches P.P. 4' to right, resting on ground plane ; 
the sides of the square are 3' and 6' respectively. 

2. Draw a pyramidal frustum same as No. 1, 3^ to left, and 
3' within P.P. 

3. Draw a triangular frustum (equilateral), sides of base 
5' and 3' respectively, height 4^ ; on G.P. 4' to right and 4' 
within P.P., vertex away, one edge parallel to P.P. 

4. Draw a square pyramidal frustum, height 4^, sides of 
square 2' and 4' respectively, standing on G.P. reversed, 
directly in front, V within P.P. 

5. Draw a conical frustum, height 5', diameters 5' and 3' 
respectively, on G.P., and touching P.P. 4' to right. 

6. Draw No. 5, 6' to left and 3' within P.P. 

7. Draw same, directly in front, 3' within P.P. and 3' 
above G.P. 

8. Draw a hexagonal frustum, height 5', sides of bases 3' 
and 2' respectively, touching P.P., resting on G.P. 4' to right. 

9. Draw same, 4' to left and 4' within P.P. 
10. Draw an octagonal frustum, height 5', edges of bases 
2' and V respectively, resting on G.P., and touching P.P. 
4' to left. 



86 



DRAWING. 



Fig. 51 shows a method of laying out a plan for a frustum 
of a cone. 




FRUSTA. 87 

Exercise XIII. 

(H = 6^ L.D. = 4^ scale ^=l^) 

1. 4' to right and 2' within P.P. draw a frustum of a 
square pyramid, edges of squares 4' and 2' respectively, height 
4'. 

2. Draw same, touching P.P. 2' to left. 

3. Draw same, directly in front, parallel to G.P, and 3' 
above it. 

4. Draw a conical frustum, height 5', diameters 3' and 2' 
respectively; frustum rests on ground, with centre of base 
3' to right and 3' within P.P. 

5. Draw a frustum of a triangular pyramid, edges of ends 
3' and 2' respectively, height 4'; it rests on G.P. with one 
edge coincident with P.P., and 3' to left. 

6. Draw a frustum of a hexagonal prism, edges of bases 
3' and 2' respectively, height 4' ; one edge of frustum is per- 
pendicular to P.P., and an angle touches it at a point, 3' to 
the right. 

7. A cone, whose height is 8' and diameter of base 4', 
touches the P.P. 4' to left; it is encircled by a rectangular 
collar whose dimensions are 4^ x 4' x P, placed centrally over 
it, 4' above the ground. The cone rests on the ground. 

A frustum of a square pyramid, whose bases are 5' and 
3' respectively, and whose height is 4', supports a cone placed 
centrally upon it ; the diameter of the cone is 3' and its height 
3'. The edge of the base touches the P.P. 2' to right. 



88 



DRAWING. 



The Sptiere. 

The perspective o£ the sphere must necessarily be repre- 
sented by a true circle, and little difficulty will be experienced 
in drawing a complete sphere. However, when a hemisphere 




THE SPHERE. 



89 



is to be represented, an apparent fallacy appears, . owing to 
the representation of the circle that shows the section of the 
sphere, A sphere must always be supposed to be drawn with 
the radius of the circle as distance. However, as the per- 
spective of the circle, viewed in any oblique position, shows 
diameters of varying length, care must be used in drawing the 
curve of the hemisphere at the greatest apparent diameter, 
and this diameter cannot be definitely determined in perspec- 
tive, if drawn in any but a direct view. The sphere rests on the 
ground at a point directly beneath the centre, and it touches 
P.P. at a point perpendicular to the vertical, from the centre. 
Example 1, — Draw a sphere, radius 2', resting on ground 
at a point 3^ to right and 2' within P.P. H-=6^ L.D. = 4', 
scale ^''=r. 

Here FA = 3', BA = AD = 4' and AC = ED = 2'. (Fig. 52. ) 
Example 2. — On centre of the top of a cube of i' edge, placed 
4' to left and touching P.P., place a sphere of radius 1|'. 




90 



DRAWING. 



Here CA = AB = AD = AE = 4^ EF = FM = 2', Gr=ll' 
hence K is centre at intersection of diagonals and HK = GF. 
(Fig. 53.) 

!NoTK — The sphere will not touch the P.P. unless HK = 
KF. 

Example S, — Draw a cube, edge 4', touching P.P. 4' to 



right, and in this place a sphere whose radius = 
sphere will touch the centre of each side. 



: 2\ Here the 




Fig. 54. 

Draw cube, and diagonals of those sides, perpendicular to 
P.P., join intersections, and bisect this horizontal line as 
shown. The circle drawn on this line as diameter will repre- 
sent the sphere, and touch the centre of each side. (Fig. 54.) 



Exercise XIY. 

(H = 6', L.D. = 4', scalei'' = l'.) 

1. Draw a sphere, radius 2', placed centrally on a cylinder 
(on end), touching P.P. 4' to left, cylinder 3' high and diame- 
ter 4'. 



THE SPHERE. 91 

2. Draw a sphere, diameter 3', 8' high, 6' to right and 6' 
within P.P. 

3. Draw a sphere, diameter 4', directly in front, touching 
P.P. 

4. Draw a sphere, radius 2', buried completely beneath the 
ground, centre of sphere 6' to right and 6^ within P.P. 

5. Place a sphere in a cubical box of 4' edge, diameter of 
sphere 4'; cube to be 4^ to left, 4' within P.P., parallel to, and 
2' above G.P. 

6. A cylinder whose height is 4' and diameter 3' stands on 
end, touching P.P. 3' to right ; this cylinder passes centrally 
through a sphere whose diameter is 4' and whose centre coin- 
cides with that of the cylinder. 



92 



DRAWING. 



JE^ore^l:i.orten.ing. 




Foreshortening consists in re- 
presenting the apparent length of 
a visible object. It depends on 
the distance of the object, and its 
position with regard to the eye. 
Thus, a lead-pencil may be so 
turned as to show only the end, 
or it may be placed so as to show 
its whole, or greatest length. 
Again, if AB represent a line of 
definite length, and O, the ob- 
server, the apparent length of 
AB as seen from O will be AD. 
(Fig. 55.) 

This representation of a line 
AB by AD, which is always less 
than AB, is called " foreshorten- 
ing." 

KoTE. — AD is always perpen- 
dicular to the lonojest side OB. 



SYNTHETIC PERSPECTIVE. 



93 



Taking an object " out of " perspective means, that when an 
object is drawn, and the position of the observer's eye given, 
the size and position of the object may be determined. 




Fig. 56. 



Here, the figure BX only, would be given, and it would be 
assumed to touch the picture plane. We first produce the 
vanishing lines DX and AC to meet in C.Y., then draw 
C.Y. S perpendicular to B.L., and make it equal to height of 
spectator; then draw H.L. parallel to B.L. through C.Y. ; 
next take a point MP at a distance to the left equal to the 
height of spectator and his distance away, combined ; then 
join MP with C, and produce it to base at E. Then, scale 
being given, find BA, AD and AC, the dimensions of the solid, 
and AS will show its distance to the left. (Pig. 56.) 



94 



DRAWING. 



Perspective Hffect. 

This consists in showing merely the appearance of an object 
when placed in a certain position. The dimensions and dis- 
tance of the object are not taken into account. 





PERSPECTIVE EFFECT. 05 

It will be remembered that if an object of less height than 
the observer, be placed on the ground, the observer will be 
able to see the upper side of it, and if placed above him he 
will see the under side ; if placed on his right, he will see 
the front and left sides ; if placed directly in front, he will 
see front side and upper or lower sides, according to the 
height of the object. 

Take a cube, for instance, placed on 
the ground parallel with the P.P. 

Now, if the cube be lower than the 
observer's eye, the upper face will be 
visible ; if raised above, the lower face, 
and so on. 

Figure 57 will illustrate perspective 
effect. 

(1) Shows object above and left of eye. 

(2) Above and directly in front of eye. 

(3) Above and to right of eye. 

(4) Level with and to left of eye. 
3 (5) Level with and directly in front. 

(6) Level with and to right. 

1 (7) Below and to left. 
^ (8) Below and directly in front. 
"I (9) Below and to right. 
I If an object, as for instance a cube, is 
I to be drawn, say to right and above the 

^ "^ — *N. '^. eye, draw first a square, then take a 

^^^ ^ ^ point to left and below, draw the van- 
ishing lines to this point, and mark off 





Cl ^^ lines for thickness, etc. 



96 DRAWING 



ANGULAE PEESPECTIYE. 

We now come to consider the rules pertaining to angular 
perspective, or the perspective of two vanishing points. If a 
rectangular object, as a cube, rests on the ground parallel to 
P.P., it is evident that its sides, if produced, will appear to 
vanish directly in front, at the point called the centre of 
vision. If we move the cube by even a small amount 
from the parallel position, its sides will no longer vanish at 
the centre of vision, but at a point to the right or left of it, 
and at a distance from it, depending on the angle which the 
sides make with the P.P. Now, in parallel perspective we 
deal with only one vanishing point — the centre of vision; but 
there are really two: for all lines parallel to the P.P., if pro- 
duced to an infinitely great distance, will appear to meet at a 
point to right or left. Hence in parallel perspective only one 
vanishing point is of practical utility. However, when the 
cube is moved out of its parallel position, this apparently- 
hidden vanishing point appears, and strikes the horizon at a 
distance from the centre of vision, depending on its angle, as 
already explained. Thus : — 

^ In Fig. 59, AB on left side shows the base of a cube in 
parallel, perspective, while in Fig. 60 AB has been moved 
around to position of AD, and AC will not now vanish to 
C.Y., but to a point Y to right of it; so also AD will not 
vanish at a point parallel to AB, but at X, a point in the 



ANGULAR PERSPECTIVE. 



97 




Fr 60. 



horizontal line to left of C.V. We wi].. now proceed to 
ascertain the positions of these points. 

Example 1. — Draw a square (side 4') lying on ground; sides 
make an angle of 45° with picture plane, and the angle 
touches the P.P. at a point 4' to the right; scale ^ =-\\ 



DRAWING. 




ANGULAR PERSPECTIVE. 99 

Here draw H.L. and B.L. as before, and take P.S. at given 
distance; then draw a straight line through P.S. parallel to 
B.L. and on each side of L.D. ; lay off the sides at required 
angle (in this case 45°); produce these lines till they meet 
the horizon in R Y.P. and L Y.P. (the vanishing points); with 
L Y.P. as a centre, and P.S. as distance, describe an arc to 
cut H.L. in R MP.; similarly find L MP. These are called 
measuring points. Now take A, 4' to right and draw A L Y.P. 
and A R Y.P.; take 4' on each side of A, namely, B and C, 
and draw B R MP and C L MP to cut vanishing lines in E 
and D. Draw D L Y.P. and E R Y.P. to cut in E. Then 
ADEE will be the square required. Eor AD^ and "f^E are 
parallel to P.S. R Y.P. placed at given angle, and AE and 
DE are parallel to P.S. L Y.P. also placed at given angle; 
and DE = AE = AB = 4^, and EE = AD = AC = 4\ (Eig. 61.) 

Example 2. — Draw a cube, edges 4', right face at an angle 
of 60° and left face at an angle of 30° with P.P. H = 6', 
L.D. = 4^, scale Y = !'• Cube to have an angle 4^ to left and 
2' within P.P. 

Draw H.L., B.L. and 00 as before; find also C.Y., R Y.P., 
L Y.P., R MP, L MP and MP (parallel perspective), as already 
shown. Take A, i' to left; join A C.Y.; erect AK = 4'; join 
K C.Y.; join also B M.P.; through E draw EL parallel to 
AK; join E RMP and E LMP, and produce them backwards 
to meet B.L. in Z and X. Mark off XC = 4; also ZD = 4'; 
join D RMP and C LMP, also L R Y.P. and LLY.P.; 
through E draw EM, and through G draw GH, parallel to 
EL; join M L Y.P. and H R Y.P. to meet at N, completing 
the required cube. Eor EM = EL = HG = AK = 4', and EG = 
EE = XC = ZD = 4'; so also MN = HL = GE, etc. (Eig. 62.) 

Note. — The point E must always be determined by parallel 
perspective ; hence necessity for finding MP ; also DP and 
Y.P. coincide at 45°, and Y.P. and MP coincide at 60°. 



100 



BRAWING. 




MISCELLANEOUS EXERCISES. 101 

Exercise XY. — Figures in Angular Perspective, 
(In the following consider H = 6^ L.D. = i', scale ^ - V.) 

1. A square whose sides are 4^, lies on ground; an angle 
touches P.P. 2' to right; angle 45°. 

2. Draw same, 3^ to left arid 2' within P.P.; right side makes 
angle of 60°. 

3. Draw same, touching P.P. directly in front; angle .45°. 

4. Draw a cube, of 4' edge, touching P.P. at a point 2' to 
right; angle 45°. 

5. Draw same, 4' to left and V within P.P.; angle 45°. 

6. Draw a square pyramid, edge of base 4', height 8', angle 
45°; touches P.P. 4^ to left. 

7. Draw same, 4' to right and 2^ within P.P.; angle 45°. 

8. Draw same, 2,' to left and 2' within P.P.; left and right 
angles, 30° and 60° respectively. 

9. Draw a triangular prism, each edge of base 3' and b' 
long; standing on end, angle touching P.P. 4' to left; angles 
60° and 60°. 

10. Draw a square pyramid, edge of base 3^, placed cen- 
trally on a cube of 4' edge; angle 45°; touches P.P. 3' to left. 
Height of pyramid 4^ 

11. Show perspective effect (angular) of a pyramid to left 
and above the eye. 

12. Show angular perspective effect of a square pyramid 
placed centrally over a cube of smaller base, to right and below 
eye. 

Miscellaneous Exercises. 

(Unless otherwise stated, consider H = 6', L.D. = 4', and 
scale i'' = l'.) 

1. Two circles, whose diameters are 4^, and intersect at 
right angles, having their common diameter perpendicular to 
ground and touching it at a point 4' to left and 4' within P.P. 

2. Draw an equilateral triangle, lying on ground plane, side 
3', vertex directed away, one side parallel to P.P.; vertex 4' 
to right and 4' within P.P. 

3. Draw a hexagon, each side 2', standing on ground, _[_ to 
P.P., one side touching it 4' to left. 



102 DRAWING. 

4. Draw a circle, diameter 4', touching G.P. and P.P., and 
perpendicular to both, 4' to right. 

5. A rod is placed obliquely in the ground, and its outside 
length is 5' ; it makes an angle of 30° with the ground and 
60° with the P.P. The rod descends toward the left, and 
lower point is 6' within P.P. and 6' to right. Draw it. 

6. Draw an octagon, side 2', lying on ground, touching 
P.P. 4' to left. 

7. Within a circle, diameter 4', lying on ground plane, 4' to 
right and 4' within, describe a square whose side shall be 
parallel to P.P. 

8. Draw a triangular prism, length 6', edges 2', parallel to 
P.P. and 3' within it, one end 4' to right, other 2' to left. 

9. Draw a cone, diameter 4', height 4', standing on ground 
plane, touching P.P. directly in front. 

10. Draw a pyramidal frustum (square), edges 2' and 4', 
height 4', touching P.P. 4^to left. 

11. Draw same, in angular perspective, angle 45°, 4' to left, 
touching P.P. 

12. Draw a sphere, diameter 4', half buried in ground, -6' 
to right and 6^ within P.P. 

13. Draw a hemisphere, plane directed towards right and 
perpendicular to P.P. and G.P., touching each ; hemisphere to 
be 4' diameter and 4^ to left. 

14. Draw a cylinder on end, diameter 4', height 4^, touch- 
ing P.P. 4' to right, and on this place a hemisphere centrally, 
4' diameter, convex surface upward. 

15. Draw a sphere touching sides of a cubical box of 4' edge, 
box on ground parallel to P.P., 4' to left and 4' within P.P. 

16. Draw a pyramid, base 4' square, 4' high, 4' to right 
and 3' within P.P. 

17. Draw an equilateral triangle, sides 3', in angular per- 
spective; angle 60°; 4' to left, 3' within P.P., on G.P. 

18. A square, sides 4', stands on ground plane perpendicular 
to it, making an angle of 45° with P.P. and 2' from it at 
nearest lower point ; it is 3' to left. 

19. Draw a triangular pyramid, height 8', each side of base 
4', presenting an angle of 60° to the P.P. 4' to left. 

20. A square prism, length 4', edge 2', stands on end, an 
angle touches P.P. directly in front; sides at 45°. This prism 



MISCELLANEOUS EXEBCISES. 103 

supports a pyramid placed evenly upon it, of equal base and 
4' in height. 

21. Draw a plinth 6' x 4' x 2^, side 6' x 4' on ground, placed 
at angles of 60° and 30°, X to left, touching P.P. 

22. Draw an ordinary Roman cross, beams 6' and 4' in 
length, and V square at ends, at an angle of 45°, 4' to right 
and 4' within P.P. 

23. Place a cube of 4' edge on top of a cylinder (on end), of 
4' diameter, centres coincident ; cylinder touches P.P. directly 
in front, 

24. Draw middle zone of a sphere whose radius is 4', height 
of zone 2', plane parallel to ground plane, centre of zone 4' to 
right, 4' wit!:in P.P. and 4' above it. H = 6', L.D. = 6', scale 

25. Draw four pyramids, each in contact at bases, 4' square 
and 4' in height, standing on ground plane at an angle of 45° 
with P.P., 4' to left and 2' within P.P. H = 6', L.D. = 6', 
scale J'' = r. 

26. Draw a frustum of a cone, height 5', diameters 3' and 
2' respectively, touching P.P. 4' to right. 

27. A cone whose slant height is 6' and diameter of base 
6', rests on ground plane, slant touching ground, parallel to 
P.P. and 6' within it, vertex directed towards left ; cone to 
be 4' to left, 

28. A cube of 4' edge contains a cylinder of equal diameter 
and height; the cube makes an angle of 45° with P.P., and 
touches P.P. 4:' to right. The cylinder is vertical. 

29. A pyramid, whose base is 4' square and whose height is 
6^, presents an angle to the P.P. 4' to left, the inclination of 
the sides being 45°. This pyramid passes centrally through a 
plinth 4' X 4' X r, placed horizontally upon it at a height of 3'. 

30. A cube, whose edges are 4', is suspended from an angle 
so as to just touch the ground directly in front, while another 
angle touches the P.P. directly in front. 

31. Draw a stove-pipe elbow, diameter of ends 6'', length of 
each half (outside) l\ The elbow rests on the ground, one 
end touching P.P. 2' to left, the other bending towards tlie 
right and parallel to P.P. H = 6', L.D. = 4', scale V=V, 

32. Show perspective effect of a pipe lying on ground paral- 
lel to P.P. and to left. 



104? DRAWING. 

33. Show same, standing on end to right and below e^e. 

34. Show angular perspective effect of a square pyramid to 
riglit, and below eye. 

35. Show perspective effect of a cylinder on end, to right 
and below. 

36. Show perspective effect of a pail with three hoops, below 
and directly in front, 

37. Show perspective effect of a water pitcher, to left and 
below; lip to right. 

38. Show perspective effect of a chair, straight back, directed 
away, angle to right and below the eye, 

39. Show hollow pipe lying on ground, perpendicular to 
P.P., to right. 

40. Show perspective effect of an ink bottle (conical), 

4L Show perspective effect of an ink bottle in form of 
pyramidal frustum, angle to left, below the eye. 

42. Show perspective effect of a plinth to left and below, 
angular. 

43. Show perspective effect of a teacup below the eye, 

44. Show perspective effect of a sphere placed centrally on 
a cube, directly in front, below the eye, 

45. Show angular perspective effect of a table below and to 
right. 

46. Show perspective effect of a triangular prism on end, 
one side perpendicular to P.P., below and to right, 

47. Show perspective effect of a reversed cone, below the- 
eye. 

48. Show perspective effect of a hexagon on one side, per- 
pendicular to ground, to right. 

49. tShow perspective effect of a hollow conic frustum lyings 
on ground parallel with P.P., larger end to right, smaller end 
to left. 

50. Show perspective effect of a door in three different 
positions, revolving round an axis through the hinges : 

(a) When shut, parallel to P.P. , 

(h) When opened at an angle of 45° with P.P. 
(c) When opened perpendicularly to P.P. 



GEOMETRICAL DEAWING. 



105 



GEOMETEICAL DEAWING. 

To construct the following figures, pupils should provide 
themselves with a pair of good compasses with pen attach- 
ment, and a ruler, with marks for inches and fractions of an 
inch. No proof is necessary, but it will be well to investi- 
gate the methods as far as possible, many of which are but 
modifications of the Euclidian. 

JVo. i. — To draw a perpendicular to a given line (a) from a 
point on the line. 




Fig. 63.- 



Fig. 64. 



Let AB be the given line, and let B be a point at which 
the perpendicular to AB is to be drawn. In first, take any 
point C above, and with distance CB describe circle, cutting 
AB in E; join EC and produce to meet circumference in D; 
join DB, which will be the perpendicular required. In second, 
take any point C above, and with B as centre and BC as dis- 
tance describe arc, cutting AB in A; then with C as centre 
and CB as distance describe arc, cutting former arc in E; 
with E as centre and same distance describe arc, cutting in D} 
8 



106 



DRAWING. 



join DB, which will be perpendicular to AB, at B. (Figs. 63 
and 64.) 

(6) From a point above or below AB, 



^ 



Fig. 65. 

Let AB be the given line and C given point above ; with 
centre C describe an arc to cut AB in D and E; with centre 
D and distance greater than half of DE describe an arc; with 
centre E and same distance describe an arc to cut former arc in 
F; join CF, which will cut AB at right angles at G. (Fig. 65.) 

iVb. ^.— To describe a square (a) on a given line. 




Fig. 66. 




In first, let AB be the given line; erect at A a perpendicular 
and make it equal to AB; with centre C and distance CA 
describe arc AD; with centre B and distance BA describe 



GEOMETRICAL DRAWING. 



107 



arc to meet former arc in D; join CD and BD, completing 
the square. (Fig. Q6,) . 

(b) On a given diagonal AB. 

With centre A and distance greater than half of AB, de- 
scribe arc CD, and with centre B and same distance describe 
an arc to cut former arc in C and D; join CD and produce 
both wajs; with centre E and distance E A or EB describe a 
circle cutting diameter in F and G; join FA, FB, GA and 
GB, completing the square. (Fig. 67.) 

iTo. 3, — To construct an oblong of given dimensions. 




Fig. 68. 

Let AB represent the greater side; erect AG perpendicular 
to AB at A, and with centre C and distance equal to AB 
describe an arc; with centre B and distance equal to AC 
describe an arc cutting former arc in D; join DC and DB, 
Gompleting the required oblong. (Fig. 68.) 

iVo. ^.— To divide a given line into (a) two equal parts. 



Fig-. GO. 



108 



DRAWING. 



Let AB be the given line ; with centre A and distance equal 
to more than half of AB describe an arc, and with centre B 
and same distance describe an arc cutting former in C and D; 
join CD, cutting AB in E into two equal parts. (Fig. 69.) 

(b) Into any number of equal parts. 

Note. — Before this can be done it is necessary to show how 
to draw a line parallel to another from an external point. 

Cy -^ D 




Fig. 70. 

Let AB be the given line and C the external point; at any 
point B and distance BC describe an arc to cut AB; with C 
as centre and CB as distance describe an arc, and with B as 
centre and distance equal to AC describe an arc cutting 
former arc in D ; join CD, which will be parallel to AB. 
(Fig. 70.) 

(b) To divide a line into any number of equal parts. 

C 




Fig. 71. 



Let AB be the given line; draw DE parallel to AB on 
either side, and on this line set off the required number of 



GEOMETRICAL DRAWING. 



109 



eqnal distances (in this case five); then join each point of 
section with A, which will divide AB into the same number 
of equal parts. (Fig. 71.) 

(c) To divide a given line proportionally to another line. 




Figr. 72. 



Let it be required to so cut AB, that the smaller part 
shall be to the greater, as the greater is to the whole line. 
Let any line CD not equal to AB, be cut in G, so that 
Ca:GD::GD:CD; place CD parallel to AB, and join CA 
and DB; produce them to meet in E; join EG to cut AB in 
F; then will AF : FB : : FB : AB. (Fig. 72.) 

1^0. 5, — To construct a triangle of given dimensions. 




Fig. 73. 

Let AB, CD and EF be the given sides, no two of which, 
taken together, are equal to, or less than, the third. Take 
one of them AB, and with centre B, and distance equal 



110 



DRAWING. 



to CD describe an arc; with centre A and distance equal to 
EF describe an arc cutting former arc in G; join GA and 
GB, completing the triangle. (Fig. 73.) 
No, 6. — To bisect a s^iven an«:le. 




Fig. 74. 



Let BAG be a given angle; with centre A and any distance 
AB describe an arc BC ; with centre B and any distance less 
than half of BC describe an arc; with centre C, and same dis- 
tance, describe an arc to cut former arc in D; join AD, which 
will bisect the angle. (Fig. 74.) 

No, 7. — To trisect a right angle. 




Fig. 75. 



Let ABC be the given right angle; with B as centre and 
at any distance BA describe the quadrant AC ; with centre A 
and distance equal to AB describe an arc to cut AC in E; and 
with centre C, and distance equal to CB or BA describe an 



GEOMETRICAL DRAWING. 



Ill 



arc to cut arc in D. Then D, E will be points of trisection, 
and lines from D and E to B will trisect the angle. (Eig, 75.) 
J\^o. 8. — To inscribe a circle in a given triangle. 




Let ABO be the given triangle; bisect the angle at A by 
AD, and the angle at B by BD, cutting AD in D ; draw DE 
perpendicular to AB at point E ; then with centre D and 
distance DE describe the circle. (Eig. 76.) 

No, 9, — To draw a circle through three given points, which, 
however, cannot be in the same straight line. 




Fig. 77. 



112 



DRAWING. 



Let A, B, C be the given points; join them to form a tri- 
angle; with centre A, and distance greater than half of AC, 
describe an arc; with centre C and same distance, describe an 
arc to cut former; join points of section of arcs; this line will 
pass through the centre of the circle; draw similar arcs on 
AB, and join points of section to meet in D, the centre; then 
a circle drawn with centre D, and distance DA, will pass 
through A, B and C respectively. (Fig. 77.) 

1^0, 10, — To find the centre of a whole or part of a circle. 




Fig. 78. 



Let BACD be a circle or arc; draw any two chords AB, 
CD, and draw arcs EF and GH, bisecting the chords, respec- 
tively; join EF and HG and produce them to meet in K; 
then K will be the centre, and if K and D be joined, KD 
will be a radius, and a circle may be thus described with it 
with the centre thus found. (Fig. 78.) 



GEOMETRICAL DRAWING. 



113 



iVo. 11. — To draw a tangent to a given circle a) from a 
point in the circumference. 




Fig. 79. 



Let C be a given point in the circumference; let A be the 
centre; join CA; at C erect the perpendicular CE, which will 
be tangent required. Also if H be the given point, join AH 
and produce it, making HG equal to part produced ; bisect 
this line by KL, which will also be a tangent at point H. If 
AC be produced, the line FC, perpendicular to the tangent at 
the point of contact C, is called a 'normal." (Fig. 79.) 

(b) To draw a tangent to a circle from an external point. 
Join point D with centre ; on DA as diameter describe a 
circle to cut given circle ; join D with points of section, which 
will form tangents with circle from external point D. 



114 



DRAWING. 



No. 12, — To construct an isosceles triangle of a given, 
altitude. 

A H 




Let AD be the given altitude ; at A and D draw perpen- 
diculars, and with A as centre and any distance describe an 
arc EF; with D as centre and any distance less than DA 
describe an arc to cut arc in E and F ; join AE and AF, and 
produce them to base, forming isosceles triangle ABC. (Fig. 
80.) 

JS'o, IS, — To construct an equilateral triangle of a given 
altitude. 

S A F 




BOG 

Fig. 81. 

Let AD be the given altitude ; through A, draw EAF per- 



GEOMETRICAL DEAWING. 115 

pendicular to AD; with centre A describe any semicircle 
EGHF, and with centre F and distance FA, describe arc to 
cut arc GH in H ; also with centre E and same distance de- 
scribe arc to cut GH in G. Join AG and AH, and produce 
them to meet BC in B and C : then ABO will be the equi- 
lateral triangle required. (Fig. 81.) 

No. 1^. — {a) To draw, from a given point in a straight line, 
an angle equal to a given angle. 




Let BAG be given angle and let the given point be at D ; 
with centre A and any distance less than a side describe an 
arc CB ; with centre D and distance equal to AB describe arc 
EF ; with centre E and distance equal to BC describe an arc 
to cut EF in F ; join DF : then the angle EDF will be equal 
to the angle BAG. (Fig. 82.) 

(h) Within a given circle to construct a triangle similar to 
another triangle. (Triangles are similar when the angles in 
one are equal to the angles in the other, each to each. They 
are similarly situated when the sides of one are parallel to the 
sides of the other, each to each. These are called homologous 
sides.) 

Let ABC be the given circle and LHK, the given triangle. 
At any point C, draw a tangent DE, and describe a semicircle 
DE ; with centre H, and any distance HM, describe an arc, 
and with centre K and same distance describe an arc. Make 
EG = NO and DF = MO ; join CF and CG, and produce tham 



116 



DRAWING. 



to the circumference in A and B ; join AB, completing the 
triangle required. (Fig. 83.) 




Fig. 83. 



iTo. 15, — To construct an equilateral triangle about a given 
circle. 




Fig. 84. 



GEOMETRICAL DRAWING. 



117 



Let BCD be given circle ; at any point A in the circumfer- 
ence, with distance equal to radius, describe an arc, cutting 
circle in B and C ; with centre C, and same distance describe 
an arc, cutting circle in D; describe similar arcs with centres 
B and D; these arcs intersect in points E, F and G; join 
these, completing the triangle required. (Fig. 84.) 

JVo. 16. — About a given circle to construct a triangle sim' 
lar to a given triangle. 




Fig. 86. 



Let DEF be the given circle and TON the given triangle ; 
find the centre G, and draw any radius GD ; at D, draw a 



118 



DRAWING. 



tangent to the circle. With centre N" and any distance MN", 
describe an arc MS, and with centre O, and same distance 
describe an arc RP ; with centre G, and distance equal to 
MN or OP, describe a circle cutting GD in H. Make arc 
HK = RP and HL = MS ; join GK and GL, and produce them 
to the circumference in E and F respectively. Draw tangents 
at E and F to meet the other tangent in A and C ; then 
triangle ABC will be similar to TON. (Figs. 85 and 86.) 

No. 17, — Within a circle to draw any number of equal 
smaller circles, each touching two others and the outer circle. 




Let KME be the given circle, and divide it (in this case) 
into six equal parts. Take centre O, and join any two, as OA, 
OB ; bisect the angle AOB by OE, and at E draw a tangent 
CD ; produce OA and OB to meet the tangent in C and D. 



GEOMETRICAL DRAWING. 119 

Bisect the angles at C and D by CF and DF, meeting at F ; 
then with centre O and distance OF describe a circle ; also 
with centre F and distance FE describe a circle : this will be 
one, and the remaining five may be similarly drawn. (Fig. 87.) 
No. 18, — To construct a regular polygon {a) on a given line. 




Fig. 88. 

Let AB be the given line, produce it both ways; then with 
centre A and distance AB, describe the senucircle DEB, 
describe also a similar semicircle AFC. Divide the circumfer- 
ence DEB into as many equal parts as the polygon is to have 
sides (in this case five), and join A with the second point of 
division ; make the arc FC = DE ; join BF and with centres 
E and F and distance EA and FB describe arcs to intersect 
at G ; join GE and GF, completing the polygon. 

Note. — This method will be clear if it be remembered that, 
if from a point within a polygon straight lines be drawn to 
the angles, the figure will be divided into as many triangles 
as it has sides, and each triangle will contain two right 
angles, but the angles around the common point within, to- 
gether make four right angles. Then if N represent the 
number of sides, the number of degrees in the angle of a 

regular polygon will be — ^^ ^, that is, ^^ \ Now, 



120 



DRAWING. 



in the above figure the line DB may be called 180°, or two 

right angles. Then the angle EAB will be represented by 

5-2 

— -— , or f of 180°; hence it is always necessary to draw 



through the second point of division. 

(h) In a given circle. 

Let ABE be the given circle; draw any diameter FC, and 
divide it into as many equal parts as the figure is to have 
sides (in this case five) ; with centre C and distance CF 
describe arc FG, and describe similar arc CG, intersecting at 
G. Draw GA from A, through second point of division ; join 
FA, and continue this around the circumference, completing 
the polygon. (Fig. 89.) 




Ifo, 19, — To construct a regular pentagon on a given line 
by a special method. 

Let AB be the given line ; describe arcs CAD and CBD, 
with radius AB ; join CD ; with centre D and distance same 
as AB describe arc EABF, cutting former arc in E and F. 



GEOMETRICAL DRAWING. 



121 



Join FG and EG, and produce them to meet arcs in H, K; 
join AH and BK; with centre H and distance HA describe 
arc; and with centre K and distance KB, describe arc cutting 
former arc in L. Join LH and LK, completing the penta- 
gon. (Eig. 90.) 




]Sfo, 20. — To construct a regular hexagon on a given straight 
line. 




Fig. 91. 



122 



DRAWING. 



Let AB be the given line ; with centre A and distance AB 
describe arc ; ^rith centre B and distance BA, describe arc 
cutting at C; join CA, CB. With centre C and distance CA, 
describe circle cutting in D, B and A ; join DC and produce 
to E; produce AC to G and BC to F; join AE, EF, FG, 
GD and DB, completing the hexagon. 

JTo. 21, — To construct a regular octagon {a) on a given 
strai^rht line. 



( 


5 H^ 




/ 


\ 


\ 




f 


/ 




> 


1 


■f 


\ 


H 


7 




c 


h 


L 


' 






Fi- 92, 



Let AB be the given line ; produce it both ways, and 
describe the semicircles CKB and AMD. Erect a perpen- 
dicular at A and at B ; bisect the right angles C AG and DBH 
by AK and BM, respectively, and erect perpendiculars KE 
and IMF, at K and ^I, respectively. With centre K and dis- 
tance KA describe arc EA cutting KE in E; draw similar 
arc BF cutting MF in F; with centre E and distance EK 
describe arc to cut AG in G ; similarly find H ; join EG, GH 
and HF, completing the octiigon. (Fig. 92.) 

(6) In a given square. 

Let BCDE be the given square; draw diagonals intersecting 
at A. With centre B, and distance BA, describe arc cutting 
sides of square in H and P ; similarly find points K, X, F, M 



GEOMETRICAL DRAWING. 



123 



and G, L; join FG, PN, ML and HK, completing the octagon. 
(Fig. 93.) 

-.0 




N M 

Fig. 93. 



No, 22, — To draw a perfect ellipse by means of the foci 
and intersecting arcs, axes being given. 




^^^ 



Fig. 94. 

Let the axes, AB and CD, be placed centrally, at right 
angles to each other ; then measure from C the distance from 
A to D, and describe arc cutting AB in F and F : these are 
the foci. Between O and F, take any number of points, 1, 2, 
etc. — the more the better ; then with centre F (left) and dis- 



124 



DRAWING. 



tance equal to distance from 1 to B, describe arcs E, E ; and 
with centre F (right) and same distance describe arcs E, E. 
Then with centre F (left) and distance equal to that from 1 
to A, describe arcs cutting the former, so also describe arcs 
from centre F (right). Thus for each point between O and F, 
we get four points. Having thus found a number of points, 
join them, or rather draw a curve through them : this curve 
will be an ellipse. (Fig. 94.) 

No, 23, — To draw an ellipse (a) by means of concentric 
circles and intersecting perpendiculars. 




Let the concentric circles EF and AB be drawn; draw 
diameters AB and CD at right angles to each other, divide 
each quadrant into the same number of equal parts, and join 
opposite points ; draw perpendiculars from the outer points 
and horizontals from the inner points to meet them ; thus 
draw perpendicular from H and horizontal from a to meet 
in 1 ; similarly find 2, 3, 4, etc., all around the circle. Draw 
a curve through the points of intersection thus found, which 
will form an ellipse. (Fig. 95.) 



GEOMETRICAL DRAWING. 125 

(6) When the major axis (transverse diameter) only is given. 




Let AB be the given diameter; divide it into four equal 
parts in G, M, D ; with centre D and distance^ DG describe 
arc EF, and with centre G and same distance describe arcs to 
intersect in E and E. Join EG, ED, EG and ED, and produce 
them to the circumference in C, H, K and L respectively. 
Then with centre E and distance EC describe arc CH, and 
with centre E and distance EK describe arc KL, completing 
the elliptical curve. (Eig. 96.) 

Note. — No part of a true ellipse is an arc of a circle. 

No. 24' — An ellipse being given, to find axis and foci. 

Draw any two parallel chords AB and CD; bisect each 
and join points of section EE, and produce each way to 
meet the circumference in G and H; bisect GH in K, and 
with centre K describe a circle to cut the ellipse in four 
points N, O, R and P; join these to form a rectangular paral- 
lelogram; bisect each side and join the opposite points of sec- 
tion, and produce both ways to meet circumference in L, M, 
T and V; then LM and TV will be the axes; and if the dis- 



126 



DRAWING. 



tance TK be taken with centre L or M, the arc will cut TV 
in S, S, which will be required foci. (Fig, 97.) 




Fig. 97. 

JTo. 25. — To draw a tangent to an ellipse (a) from a point 
in the circumference. 




Fig. 98. 



GEOMETRICAL DRAWING. 



127 



Let B, C be the foci and A, given point on the curve; join 
BA and CA and produce them to D and F; bisect the angle 
BAD by AK, and the angle DAF by AG. Then will KA be 
a tangent and GA a perpendicular or normal to it, at the 
point of contact, A. (Fig. 98.) 

(b) From an external point. 




Fig. 99. 



Let A be an external point; draw major axis BC, and on 
it describe the semi-circle BFC; draw a tangent AF to the 
circle at F; draw FE perpendicular to BC, cutting curve in 
G; join AG, which will be a tangent, to the curve. (Fig. 99.) 

N'o, 26. — To draw an oval of a given width. 

Let AB be the given width ; bisect it in C, and on it describe 
the circle ADB; draw CD at right angles to AB; with centre 
B and distance BA describe curve AE; similarly describe 



us 



DRAWING. 



curve BI ; join BD and AD, and produce them to meet curve 
in E and F ; with centre J) and distance DE describe curve 
EE, completing the oval. (Fig. 100.) 




Jfo, 27, — To construct the involute of the circle. 

Divide the circle into any number of equal parts and draw 
the radii, numbering them 1, 2, etc. Draw the tangents, mak- 
ing the first the length from 1 to 2, the second twice this 
length, the third three times, and so on. When all the tan- 
gents have been drawn thus, begin again at 1 by producing 
it, and so get a second series of points. Then draw a curve 
through the points, commencing with 8, or last, and joining 
it with 1, then 2, etc. (Fig. 101.) 



GEOMETRICAL DRAWING. 



129 



7 


f\ 


/t\' 


5 / 


3^ 


\ 


5 . 


\y 



3-. 



Fig:. 101. 



]}^o. 28, — To find {a) a mean proportional between two given 
lines. 




Fig. 102. 



Let AC and BC be the given lines ; place them in a 
straight line AB ; bisect AB in D, and on AB describe 



130 



DRAWING. 



semicircle AEB ; through C draw CE at right angles to AB ' 
then will AC : CE : : CE : CB. (Fig. 102.) 

(b) To draw a third proportional to two given lines 




Let AB and BC be the given lines ; draw AE, making any 
angle with AC ; make AD = BC ; join BD, and through C 
draw CE parallel to BD, meeting AD produced : then will 
AB:BC::BC:DE. (Fig. 103.) 

If BC be greater than AB, DE will be a third proportional 
greater ; but if BC be less than AB, DE will be less 

IS 




Fig. 104. 



GEOMETRICAL DRAWING. 



131 



No. 29, — To draw a circle of given radius which shall touch 
both lines of a given angle. 

Let B AC be the given angle and ST the given radius ; bisect 
angle by line AF; erect on either line a perpendicular DE 
equal to ST; through E draw EG parallel to AD, and cutting 
AF in G ; then a circle drawn with centre G and radius equal 
to DE or ST, will touch the sides AB and AD. (Fig, 104.) 

iTo. SO. — To draw a circle of given radius which shall touch 
another given circle and a given straight line. 




Fig. 105. 



Let FG be a given line, BX a given circle, and DE a given 
radius; draw a line GH perpendicular to FG and equal to 
DE; through H draw HK parallel to FG; draw any radius 
AB and produce it, making the part produced equal to DE; 
then with centre A and distance AC, describe an arc cutting 
HK in K : the circle drawn with centre K, and at a distance 
equal to GH or DE will touch the circle BX and the line 
FG. (Fig. 105.) 



132 DRAWING. 



Grraded E:?cerci^e«. 

1. Construct a square whose side is V , * 

2. Construct a square whose diagonal is M' , 

3. Construct an oblong whose sides are \^' and 2|'' respec- 
tively. 

4. A rectangular field is 900 yds. long and 400 yds. wide; 
divide it into four equal fields each 400 yds. long. Scale 
100 yds. = 1". 

5. Divide a line Z^ long into two parts in the ratio 3 : 4. 

6. Draw a line parallel to and between two other parallel 
lines 2^'' apart, the line to be twice as near one as the other. 

7. Construct an equilateral triangle whose side is ^" , 

8. Centrally within the triangle in Ko. 7 construct a tri- 
angle whose side is \Y - 

9. Construct a triangle whose sides are 2|-^^ 3;^'^ and 3|'' 
respectively. 

10. Inscribe a circle in a triangle whose sides are same as 
No. 9. 

11. The diagonal of a parallelogram is M and one side is 2'. 
Draw it. 

12. Find the extent of an angle of 22l°,of 371°, of 41 J°. 

13. Construct an isosceles triangle whose base is V and 
vertical angle 37 1°. 

14. Construct an isosceles triangle whose base is 2'' and 
altitude 3^^ 

15. Construct an equilateral triangle about a circle whose 
diameter is M' . 

16. Construct a triangle whose sides are in the ratio 3:4:5 
about a circle whose diameter is M' , 

17. Describe a circle about a square whose side is Z'\ 

18. Within a triangle whose sides are 3^', M' and ^" respec- 
tively, inscribe a circle. ^ 

19. Construct a regular pentagon whose side is 1^'^ 

20. Construct a regular heptagon in a circle M diameter. 



GRADED EXERCISES. 133 

21. Draw a hexagon whose side is V , 

22. Draw a hexagon within an equih^^teral triangle of 3''. 

23. Construct a regular octagon whose side is \^\ 

24. Construct a regular octagon in a square, side 3''. 

25. Construct a regular octagon in a circle of 3^' diameter. 

26. Inscribe seven equal circles in a circle of Z" diameter. 

27. The diameters of an ellipse being ?)^' and2 ^ respec- 
tively, draw it. 

2'^, Draw an elliptical curve on a transverse axis of 3'^ 

29. Draw an elliptical figure from two squares, diagonal of 
each 3''. 

30. A circle and an ellipse touch the angles of an oblong 
2>" y. 5'', find axes and foci of the ellipse. 

31. Draw an oval whose shorter axis is ^Y - 

32. Construct an involute to a circle whose diameter is y , 

33. Find a mean proportional between two lines 2^ and 
3|'' respectively. 

34. Draw a third proportional (greater) to two lines 2^' 
and Zy respectively. 

35. Draw same as No. 34, less. 

36. Lay out a circular garden whose radius is 30 yds., which 
shall just touch a fence on one side, and another garden whose 
radius is 50 yds. on the other. Scale 20 yds. = 1''. 

37. Lay out a circular garden, radius 30 yds., which shall 
touch two fences not parallel. Scale 20 yds. = 1^^ 

38. Lay out a circular garden, radius 30 yds., which shall 
touch two fences not parallel, and whose edge shall just touch 
a tree in a given position. Scale 20 yds. = \" , 

39. The axes of an elliptical flower garden are 40 and 60 
yds. respectively, and a point is taken 60 yds. to left of the 
shorter (produced), and 40 yds. above the longer (produced). 
Draw a path from this point to touch the elliptical garden. 
Scale 20 yds. = I'''. Longer axis horizontal. 

40. Construct a triangle whose angles are 75°, 45° and 60° 
respectively, on a line 2" in length. 

41. Two circles, whose diameters are each 3|'', intersect, the 
intercepted arc of each being one-fifth of the whole circum- 
ference. 

42. The base of a right-angle triani]:lc being 2\ and the 



134 DRAWING. 

perpendicular on the hypothenuse from the right angle being 
i^", construct the triangle. 

43. Two lines meet at a point. Find a point between them 
that will be 2'' from one and 3'^ from the other. 

44. Show that the number of degrees in the angle of any- 
regular polygon is represented by '^ \ where n = num- 
ber of sides. 

45. Show that a circle is but a particular form of an 
ellipse. 

46. If an ellipse be drawn with a string around two fixed 
pins as foci, show that the sum of the distances from any 
point in the curve to the foci is the same. 

47. From the point found in 'No. 43 draw two equal 
straight lines to the given line. 

48. Draw a parallelogram 3^ x 5^'\ and within this draw 
one of half the size similar and similarly situated. 

49. If one hexagon be inscribed in, and another inscribed 
about, a circle, show that their areas are in ratio 3 : 4. 

50. Three circular gardens, diameters 20 yds., 30 yds. and 
40 yds. respectively, are to be placed with walks of 5 yds. 
between them. Scale 20 yds. = 1''. 

51. Divide a triangle whose sides are 3^', V and 5'' respec- 
tively, into four equal and similar triangles. 

52. From the vertex of a scalene triangle draw a straight 
line to the base which shall exceed the less side as much as it 
is exceeded by the greater. 

53. One of the acute angles of a right-angled triangle is 
three times as great as the other; trisect the smaller. 

54. One side of a right-angled triangle is V, and the differ- 
ence between the hypothenuse and the sum of the other two 
sides is 2" ; construct the triangle. 

55. The altitude of an equilateral triangle is 2'^; construct it. 
5Q. Place a straight line, 3'^ in length, between two straight 

lines, each 2^' in length, which meet, so that it shall be equally 
inclined to each of them. 

57. Describe an isosceles triangle upon a given base, having 
each of the sides double of the base. 

58. Draw a square equal in area to two unequal oblongs. 



GRADED EXERCISES. 135 

59. Given the base, the vertical angle and the perpendicular 
of a plane triangle, construct it. 

60. Cut off two-thirds of an isosceles triangle by a line 
parallel to the base. 

61. Describe two circles with given radii which shall cut 
.each other and have the line between the points of section 
equal to a given, limited line. 

62. Describe a circle with a given centre cutting a given 
circle in the extremities of the diameter. 

63. Describe a circle which shall pass through a given point 
and which shall touch a given straight line in a given point. 

64. Describe a circle which shall touch a given straight 
line at a given point and bisect the circumference of a given 
circle. 

65. Two circles are described about the same centre; draw 
a chord to the outer circle, which shall be divided into three 
equal parts by the inner one. What are the limits of the 
diameters *? 

66. The perimeter of an oblong is 20'^, and the sides are in 
ratio 3:2; construct it. 

67. Construct an isosceles triangle of given vertical angle 
and given altitude. 

68. Draw a square equal in area to an oblong 3'^ x V » 

69. Describe a circle of given radius to touch two points. 
What limits the position of the points % 

70. Show how to draw a similar triangle within another. 

71. Two equal ellipses, axes ^" and Z\ cut each other at 
right angles, and their centres are coincident ; draw them. 

72. Trisect a given line. 

73. Show how an angle may be trisected (mechanically). 

74. Divide a square into three equal parts by lines drawn 
from an angle. 

75. Divide an oblong into three equal parts by lines drawn 
from an ano^le. 

76. Find the number of degrees in the angle of a regular 
duodecagon. 



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